A New Operator Theory Similar to Pseudo-Differential Operators

††footnotetext: † Corresponding author. Yan’an Second School, Yan’an 71600,
China
E-mail<EMAIL_ADDRESS>
b) School of Electronic and Information Engineering, BUAA, Beijing 100191,
China
E-mail<EMAIL_ADDRESS>
Guangqing Bi a)†, Yuekai Bi b)

###### Abstract

We summarize and extend the correlative definitions and principles of abstract
operators, discuss the relation between abstract operators and pseudo-
differential operators, add several new algorithms, furthermore, develop the
theory of partial differential equations with abstract operators, and then
systematically expound the basic methods. By combining abstract operators with
the Laplace transform, we can easily derive the explicit solution of initial
value problem of linear higher-order partial differential equations for
n-dimensional space, and establish the general theory of linear higher-order
partial differential equations.

Keywords: Abstract operators, Symbol, Partial differential equations, Laplace
transform, Pseudo-differential operators

MSC(2010) Subject Classification: 35G10, 44A10, 47A56, 35S05, 35G05

##  1 Introduction

In 1960s, the general theory of linear partial differential equations made
important progress by using the generalized function and its Fourier
transform, further, the theory of pseudo-differential operators. The
significance of the latter theory lies in introducing the concept of symbol,
but due to the lack of awareness of the analytic continuous fundamental
theorem (See [1]), the concept of abstract operators remained unclear. Without
the theories of abstract operators, the operator method of differential
equations is just a technique, and the solution of partial differential
equations is normally complex. In 1997, the author delivered the analytic
continuous fundamental theorem, based on which the author introduced the
concept of abstract operators, and derived the algorithms of five types of
abstract operators (See [1]) such as

$\exp(h\partial_{x}),\;\sin(h\partial_{x}),\;\cos(h\partial_{x}),\;\sinh(h\partial_{x})\;\mbox{and}\;\cosh(h\partial_{x}).$

Where $\cos(ih\partial_{x})=\cosh(h\partial_{x})$,
$\sin(ih\partial_{x})=i\sinh(h\partial_{x})$. Obviously, the abstract operator
uses the same notation as infinite order differential operator, as it is the
extension of the latter. However, within the frame of infinite order
differential operators, complex operators are difficult to define, such as

$\exp(tP(\partial_{x})),\;\cos(tP(\partial_{x})^{1/2}),\;\frac{\sin(tP(\partial_{x})^{1/2})}{P(\partial_{x})^{1/2}},\;\cosh(tP(\partial_{x})^{1/2}),\;\frac{\sinh(tP(\partial_{x})^{1/2})}{P(\partial_{x})^{1/2}}.$

But as abstract operators, we can easily establish their algorithms (See [1],
[2]), therefore the general solving procedure of initial value problem of
linear partial differential equations with constant coefficients is derived
[1], [2], [3], this approach can also be applied in solving the partial
differential equations with variable coefficients containing $t$.

According to [4], pseudo-differential operators are defined as:

If $a(x,\xi)\in{C}^{\infty}(\mathbb{R}_{x}^{n}\times\mathbb{R}_{\xi}^{n})$,
and for arbitrary $\alpha,\beta\in\mathbb{N}^{n}$ and an real number $m$,

$|\partial_{\xi}^{\alpha}\partial_{x}^{\beta}{a}(x,\xi)|\leq{C}_{\alpha,\beta}(1+|\xi|)^{m-|\alpha|}$

is tenable, where $C_{\alpha,\beta}$ is a constant, then the linear continuous
mapping $A$ of
$\mathscr{S}(\mathbb{R}^{n})\rightarrow\mathscr{S}(\mathbb{R}^{n})$ can be
defined as:

$Au(x)=(2\pi)^{-n}\int{e}^{i\xi{x}}a(x,\xi)\hat{u}(\xi)d\xi,$ (1)

which are called the pseudo-differential operators, denoted by $a(x,D)$, where
$a(x,\xi)$ are the symbols of $a(x,D)$, $D$ denotes

$D^{\alpha}=\left(\frac{1}{i}\partial_{x_{1}}\right)^{\alpha_{1}}\left(\frac{1}{i}\partial_{x_{2}}\right)^{\alpha_{2}}\cdots\left(\frac{1}{i}\partial_{x_{n}}\right)^{\alpha_{n}}.$

Obviously, this definition means the following equation is tenable, namely

$a(x,D)e^{i\xi{x}}=a(x,\xi)e^{i\xi{x}}.$ (2)

Therefore pseudo-differential operators are similar to abstract operators. As
the definition formula of pseudo-differential operators is a Fourier integral
expression, their symbols are largely restricted to ensure the convergence of
Fourier integrals. But for many specific problems, the symbols of pseudo-
differential operators should not be restricted that much. Thus, a better
definition of symbols can be expressed as:

$\sigma_{A}(x,\xi)=e^{-i\xi{x}}Ae^{i\xi{x}}.$ (3)

Where $A$ are properly extended pseudo-differential operators. Certainly, this
extension still has its limitations as it involves another type of Fourier
integrals.

Then can we get rid of Fourier integrals when defining pseudo-differential
operators? Let us study the following expression:

$Ae^{i\xi{x}}=\sigma_{A}(x,\xi)e^{i\xi{x}}.$ (4)

Obviously, the properties of $A$ depend on $\sigma_{A}(x,\xi)$. (4) indicates
the mapping relation between $A$ and $\sigma_{A}(x,\xi)$. For a given function
$\sigma_{A}(x,\xi)$, if the corresponding algorithms of $A$ can be derived
from this type of mapping relations, to determine the domain and range of $A$,
then this type of mapping relations can be the best definition of pseudo-
differential operators. Luckily, this idea totally works by using the analytic
continuous fundamental theorem or the abstract operator fundamental theorem.
The pseudo-differential operators established by using this method are the
abstract operators.

Now let us carry out a systematic review of the basic methods of abstract
operators, and prove that if we combine abstract operators with the Laplace
transform on $C^{\infty}(\mathbb{R}_{x}^{n}\times\mathbb{R}_{t}^{1})$, then no
matter how complex the initial value problem of linear partial differential
equations is, the solving process is simply easy. In addition, in order to
achieve self consistency theoretically, it is necessary to derive rules of
differentiation from the view of abstract operators.

##  2 Basic theory of abstract operators

###  2.1 Basic concepts, research method and basic operations

Definition 1. (See [1]) Within each convergence circle of analytic functions,
if the effects on the series term by term from the linear operator converge
uniformly to the effects on the sum function, then the operator is called
having the analytic continuity.

Definition 2. The linear continuous mapping $A$ of
$\mathscr{S}(\mathbb{R}^{n})\rightarrow\mathscr{S}(\mathbb{R}^{n})$ has the
Fourier continuity, if $A$ acts on an arbitrary function
$f\in\mathscr{S}(\mathbb{R}^{n})$, we have

$Af(x)=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}\hat{f}(\xi)\,Ae^{i\xi{x}}\,d\xi.$

Where the Fourier transform $\hat{f}=\mathscr{F}f$ is defined as

$\hat{f}(\xi)=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}f(x)e^{-i\xi{x}}\,dx,\quad\forall{f}\in\mathscr{S}(\mathbb{R}^{n}).$

The Fourier transform $\mathscr{F}$ is an isomorphic mapping in the space
$\mathscr{S}(\mathbb{R}^{n})$, and its inverse transform is

$\mathscr{F}^{-1}\hat{f}(\xi)=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}\hat{f}(\xi)e^{i\xi{x}}\,d\xi=f(x).$

Definition 3. (See [1]) $x^{\alpha}$, $x\in{\mathbb{R}^{n}}$ is called the
base function, its exponential form is $e^{\xi{x}}$, where
$\alpha\in\mathbb{N}^{n},\;\xi\in{\mathbb{R}_{n}}$ or $\mathbb{C}^{n}$ are
respectively called the character of $x^{\alpha}$ and $e^{\xi{x}}$.

Definition 4. (See [1]) A class of linear operators is called the abstract
operators, denoted by
$f(t,\partial_{x}),\,x\in\mathbb{R}^{n},\,t\in\mathbb{R}^{1}$, and if it acts
on the exponential base function $e^{\xi{x}}$, we have

$f(t,\partial_{x})e^{\xi{x}}=f(t,\xi)e^{\xi{x}},\qquad\forall{f(t,\xi)}\in{C^{\infty}}(\mathbb{R}_{t}^{1}\times\mathbb{R}_{\xi}^{n}).$

Which is called the abstract operators taking $\partial_{x}$ as the operator
element, where $f(t,\xi)$ is called the symbol of the abstract operators
$f(t,\partial_{x})$.

If abstract operators $f(t,\partial_{x})$ is a linear operator having the
Fourier continuity, then it similar to a pseudo-differential operators.

Definition 5. A class of linear operators is called the abstract operators,
denoted by $f\left(t,x\frac{\partial}{\partial{x}}\right)$,
$x\in\mathbb{R}^{n},\,t\in\mathbb{R}^{1}$, where

$x\frac{\partial}{\partial{x}}=\left(x_{1}\frac{\partial}{\partial{x}_{1}},\;x_{2}\frac{\partial}{\partial{x}_{2}},\;\ldots,\;x_{n}\frac{\partial}{\partial{x}_{n}}\right),$

and if it acts on the base function $x^{\alpha}$, we have

$f\left(t,\,x\frac{\partial}{\partial{x}}\right)x^{\alpha}=f(t,\alpha)x^{\alpha},\qquad\forall{f(t,\alpha)}\in{C^{\infty}}(\mathbb{R}_{t}^{1}\times\mathbb{N}^{n}).$

Which is called the abstract operators taking $x\frac{\partial}{\partial{x}}$
as the operator element, where $f(t,\alpha)$ is called the symbol of the
abstract operators $f\left(t,x\frac{\partial}{\partial{x}}\right)$.

Definition 6. We call any operator element and common variables the abstract
element, denoted by $X,Y,\ldots$ and $X=(X_{1},X_{2},\ldots,X_{n})$,
$Y=(Y_{1},Y_{2},\ldots,Y_{n})$, etc.; Accordingly, $f(X)$, $f(Y)$, etc.,
represent the abstract operators or common functions, which is called the
operator function.

Abstract operators fundamental theorem. Let $A,B,\;A^{\prime},B^{\prime}$ be
the abstract operators, if there are functions
$v(t)\in\mathscr{D}(A^{\prime}),\;u(t)\in\mathscr{D}(B^{\prime}),\;t\in\mathbb{R}^{1}$
and abstract elements $X,Y$, making one of the following two expressions
tenable

$AX^{\alpha}{A}^{\prime}v(t)=BY^{\alpha}{B}^{\prime}u(t)\quad\mbox{or}\quad
Ae^{\xi{X}}{A}^{\prime}v(t)=Be^{\xi{Y}}{B}^{\prime}u(t),$

and the expressions of $A,B,\;A^{\prime},B^{\prime}$ do not explicitly contain
the character $\alpha\in\mathbb{N}^{n}$,
$\xi\in\mathbb{R}_{n}\,\mbox{or}\,\mathbb{C}_{n}$, then

$Af(X)A^{\prime}v(t)=Bf(Y)B^{\prime}u(t),\qquad\forall{f(X),f(Y)}\in\mathscr{D}(A)\cap\mathscr{D}(B).$

Where $\mathscr{D}(\cdot),\,\mathscr{R}(\cdot)$ are domains and ranges of
operators.
$\mathscr{D}(f(X))=\mathscr{R}(A^{\prime}),\;\mathscr{D}(f(Y))=\mathscr{R}(B^{\prime})$,
$\mathscr{D}(A)=\mathscr{R}(f(X)),\;\mathscr{D}(B)=\mathscr{R}(f(Y))$.

When $A^{\prime}v(t)=B^{\prime}u(t)=I$, and
$X=x\in\mathbb{R}^{n},\;Y=y\in\mathbb{R}^{n}$, this theorem becomes the
following Analytic continuous fundamental theorem:

If there are functions $x(t),y(t)\in\mathbb{R}^{n},\;t\in\mathbb{R}^{1}$,
making one of the following two expressions tenable

$Ax^{\alpha}=By^{\alpha}\quad\mbox{or}\quad Ae^{\xi{x}}=Be^{\xi{y}},$

and the expressions of abstract operators $A,B$ do not explicitly contain the
character $\alpha\in\mathbb{N}^{n}$,
$\xi\in\mathbb{R}_{n}\,\mbox{or}\,\mathbb{C}_{n}$, then

$Af(x)=Bf(y),\qquad\forall f(x),f(y)\in\mathscr{D}(A)\cap\mathscr{D}(B).$

Especially, when $A,B$ are the linear operators having the analytic
continuity, the analytic continuous fundamental theorem has been proved by G.Q
Bi in 1997 (See [1]).

The abstract operators fundamental theorem is the fundamental property of
abstract operators, which directly extends the domains of abstract operators
from power functions and exponential functions to a wider range of functions.

Definition 7. A linear operator equation, if it’s true for arbitrary functions
or operator functions in a certain range, then it is called an operator
formula. Particularly, the operator formulas that determine the domain and
range of abstract operators are called the algorithm of the abstract
operators.

Definition 8. The relational expression between each component of the
character $\alpha=(\alpha_{1},\alpha_{2},\ldots,\alpha_{n})$ or
$\xi=(\xi_{1},\xi_{2},\ldots,\xi_{n})$ is called the characteristic equation.
In general, an operator formula becomes the corresponding characteristic
equation, when where arbitrary function of which is the base function.

Corollary 1. As the base function can be expressed by both power function and
exponential function, a single operator formula can be corresponding to two
different characteristic equations, conversely, a single characteristic
equation can also be corresponding to two different operator formulas.

Constructing the operator equation tenable for the base function by the
characteristic equation through the definition of abstract operators, and then
deducing that it is also tenable for arbitrary functions or operator functions
in a certain range by using the analytic continuous fundamental theorem or the
abstract operators fundamental theorem, thus we derive new operator formulas.
Finding or establishing a new operator formula requires the knowledge of the
corresponding characteristic equation in advance, without knowing the specific
form of the new operator formula. Therefore, it all boils down to seek or
construct appropriate characteristic equations. The key to transform the
characteristic equation to the corresponding operator formula, is constructing
the operator equation true for the base function by using the specific form of
the characteristic equation and the definition of abstract operators, and
also, only when the operator constructed is linear and the expression of the
linear operator doesn’t explicitly contain the character of the base function
can we derive that the operator equation is not only true for the base
function, but also for arbitrary functions in a certain range, according to
the analytic continuous fundamental theorem or the abstract operators
fundamental theorem. This is the research method of abstract operators.

Example 1. Let $x\in\mathbb{R}^{n},\;\xi\in\Omega\subseteq\mathbb{R}_{n}$,
then $\forall{f}(\xi)\in{C}^{\infty}(\Omega)$ we have

$\displaystyle f(\partial_{x})\sin{bx}$ $\displaystyle=$
$\displaystyle\left.\sin\left(bx+\frac{\pi}{2}\,\xi_{1}\frac{\partial}{\partial\xi_{1}}+\frac{\pi}{2}\,\xi_{2}\frac{\partial}{\partial\xi_{2}}+\cdots+\frac{\pi}{2}\,\xi_{n}\frac{\partial}{\partial\xi_{n}}\right)f(\xi)\right|_{\xi=b},$
$\displaystyle f(\partial_{x})\cos{bx}$ $\displaystyle=$
$\displaystyle\left.\cos\left(bx+\frac{\pi}{2}\,\xi_{1}\frac{\partial}{\partial\xi_{1}}+\frac{\pi}{2}\,\xi_{2}\frac{\partial}{\partial\xi_{2}}+\cdots+\frac{\pi}{2}\,\xi_{n}\frac{\partial}{\partial\xi_{n}}\right)f(\xi)\right|_{\xi=b}.$

Where $bx=b_{1}x_{1}+b_{2}x_{2}+\cdots+b_{n}x_{n}$.

Proof. Selecting the following two expressions from the known formulas as the
characteristic equations, namely

$\partial_{x}^{\alpha}\sin{bx}=b^{\alpha}\sin\left(bx+\frac{\pi}{2}|\alpha|\right)\quad\mbox{and}\quad\partial_{x}^{\alpha}\cos{bx}=b^{\alpha}\cos\left(bx+\frac{\pi}{2}|\alpha|\right).$

Taking $\alpha\in\mathbb{N}^{n}$ as the character of the base function,
$\partial_{x}$ and $\xi\frac{\partial}{\partial\xi}$ as the operator elements,
we have the characteristic equation with the following two operator equations
by Definition 5:

$\partial_{x}^{\alpha}\sin{bx}=\left.\sin\left(bx+\frac{\pi}{2}\,\xi_{1}\frac{\partial}{\partial\xi_{1}}+\frac{\pi}{2}\,\xi_{2}\frac{\partial}{\partial\xi_{2}}+\cdots+\frac{\pi}{2}\,\xi_{n}\frac{\partial}{\partial\xi_{n}}\right)\xi^{\alpha}\right|_{\xi=b},$
$\partial_{x}^{\alpha}\cos{bx}=\left.\cos\left(bx+\frac{\pi}{2}\,\xi_{1}\frac{\partial}{\partial\xi_{1}}+\frac{\pi}{2}\,\xi_{2}\frac{\partial}{\partial\xi_{2}}+\cdots+\frac{\pi}{2}\,\xi_{n}\frac{\partial}{\partial\xi_{n}}\right)\xi^{\alpha}\right|_{\xi=b}.$

According to the abstract operators fundamental theorem, we get the Example 1.

Definition 9. Let $A$ be a linear operator having the analytic continuity or
Fourier continuity, if there is another linear operator, denoted by $A^{-1}$,
making $AA^{-1}=A^{-1}A=I$, then $A^{-1}$ is called the inverse operator of
$A$.

By Definition 4 we have the following Corollary:

Corollary 2. (See [1]) The operator algebras formed by all the abstract
operators $f(t,\partial_{x})$, are isomorphic to the algebras formed by their
symbols $f(t,\xi)$. This isomorphism is determined by
$f(t,\partial_{x})\leftrightarrow{f}(t,\xi)$, and

$f(t,\partial_{x})\pm{g}(t,\partial_{x})\leftrightarrow{f}(t,\xi)\pm{g}(t,\xi),\quad{f}(t,\partial_{x})\circ{g}(t,\partial_{x})\leftrightarrow{f}(t,\xi)g(t,\xi).$

Especially, the abstract operators $f(\partial_{x})$ and
$g(\partial_{x}),\;x\in\mathbb{R}^{n}$ are each other’s inverse operator, if
and only if their symbols $f(\xi)$ and $g(\xi)$ satisfy
$f(\xi)g(\xi)=1,\;\xi\in\mathbb{R}_{n}$.

For instance,
$\partial_{x_{i}}\sin(x\partial_{y})=\partial_{y_{i}}\cos(x\partial_{y})$,
$\partial_{x_{i}}\cos(x\partial_{y})=-\partial_{y_{i}}\sin(x\partial_{y})$,
where $x,y\in\mathbb{R}^{n}$.

Corollary 3. (See [1]) Let $g(x)\in{C^{\infty}}(\mathbb{R}^{n})$, if
$g(\partial_{\xi})(e^{\xi{x}}f(\xi))$ is continuous at $\xi=a$, then

$f(\partial_{x})(e^{ax}g(x))=g(\partial_{\xi})(e^{\xi{x}}f(\xi))|_{\xi=a}.$
(6)

Obviously, by Definition 5 we can also get similar Corollary.

Example 2. If function $f(x),x\in\mathbb{R}^{n}$ satisfies

$\int_{\mathbb{R}^{n}}f(\eta)\exp\left(-\frac{|\eta-2\alpha{x}|^{2}}{4\alpha}\right)d\eta<+\infty,\quad\eta\in\mathbb{R}^{n},\;\alpha>0,$

then

$f(\partial_{x})e^{\alpha|x|^{2}}=e^{\alpha|x|^{2}}\frac{1}{2^{n}(\alpha\pi)^{n/2}}\int_{\mathbb{R}^{n}}f(\eta)\exp\left(-\frac{|\eta-2\alpha{x}|^{2}}{4\alpha}\right)d\eta.$
(7)

Proof. In the Corollary 3, let
$g(x)=e^{\alpha|x|^{2}}\in{C^{\infty}}(\mathbb{R}^{n})$ and $a=0$, we have

$\displaystyle f(\partial_{x})e^{\alpha|x|^{2}}$ $\displaystyle=$
$\displaystyle\left.\exp(\alpha|\partial_{\xi}|^{2})(e^{\xi{x}}f(\xi))\right|_{\xi=0}=\left.\exp(\alpha|x+\partial_{\xi}|^{2})f(\xi)\right|_{\xi=0}$
$\displaystyle=$ $\displaystyle
e^{\alpha|x|^{2}}\left.\exp(2\alpha{x}\partial_{\xi})\exp(\alpha|\partial_{\xi}|^{2})f(\xi)\right|_{\xi=0}$
$\displaystyle=$ $\displaystyle
e^{\alpha|x|^{2}}\frac{1}{2^{n}(\alpha\pi)^{n/2}}\left.\exp(2\alpha{x}\partial_{\xi})\int_{\mathbb{R}^{n}}f(\eta)\exp\left(-\frac{|\eta-\xi|^{2}}{4\alpha}\right)d\eta\right|_{\xi=0}$
$\displaystyle=$ $\displaystyle
e^{\alpha|x|^{2}}\frac{1}{2^{n}(\alpha\pi)^{n/2}}\left.\int_{\mathbb{R}^{n}}f(\eta)\exp\left(-\frac{|\eta-\xi-2\alpha{x}|^{2}}{4\alpha}\right)d\eta\right|_{\xi=0},\;\alpha>0.$

Where
$x\partial_{\xi}=x_{1}\partial_{\xi_{1}}+x_{2}\partial_{\xi_{2}}+\cdots+x_{n}\partial_{\xi_{n}}$,
$|x|^{2}=x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{n}$.

Clearly, by (7) the $f(\eta)$ can also be a bounded measurable function.
Therefore, Corollary 3 is an important way to extend the symbols and domains
of abstract operators. In fact, limitations on the symbol $f(t,\xi)$ and the
operator element $X$ will be broadened, and more general abstract operators
will be introduced to deal with problems involved in our further discussion.

There are two major research methods of the operator theory. One determines
the domain and range of an operator first, then derives its operator algebra.
The other abstractly determines an operator algebra, then derives the domain
and range of the operator. The former is applied by pseudo-differential
operators, while the latter is applied by abstract operators. The definition
of abstract operators actually determines a kind of excellent operator
algebras, then establishes different algorithms for different operators using
the abstract operator fundamental theorem, further, derives different domains
and ranges for abstract operators with different symbols. In other words, the
domain and range of an abstract operator depend on the algorithm it applies.
Therefore, a significant feature of abstract operators is focusing on the
research of algorithms, without depending on the Fourier integral transform.
The basic theory of abstract operators is a theory about algorithms, which has
already become an important tool to seek and calculate explicit solutions of
linear higher-order partial differential equations.

###  2.2 Algorithms of Abstract operators on differentiable function

In order to achieve theoretical rigor, firstly let us derive the rules of
differentiation from the view of abstract operators, making the concept of
ordinary or partial differential operator on the basis of the abstract
operators itself.

Differential operators can be defined as the following abstract operators,
namely

$x\frac{d}{dx}\,x^{\alpha}=\alpha{x}^{\alpha},\quad\frac{d}{dx}\,e^{\xi{x}}=\xi{e}^{\xi{x}},\qquad\forall{x}\in\mathbb{R}^{1}.$

Where $\alpha\in\mathbb{N}^{1},\;\xi\in\mathbb{R}_{1}$ are the symbols of
abstract operators $x\frac{d}{dx}$ and $\frac{d}{dx}$ respectively.

Let $a$ and $b$ be the characters of base functions, as the characteristic
equation, the binomial formula of integer power can be expressed as

$(a+b)^{n}=\sum^{n}_{j=0}{n\choose{j}}a^{j}b^{n-j},\quad\forall{n}\in\mathbb{N}.$

By base functions $e^{ax}$ and $e^{bx}$ we can combine the binomial formula
with the following operator equations:

$\frac{d^{n}}{dx^{n}}\left(e^{ax}\cdot{e}^{bx}\right)=\sum^{n}_{j=0}{n\choose{j}}\frac{d^{j}}{dx^{j}}e^{ax}\cdot\frac{d^{n-j}}{dx^{n-j}}e^{bx}.$

According to the analytic continuous fundamental theorem, it is tenable,
namely

$\frac{d^{n}}{dx^{n}}(vu)=\sum^{n}_{j=0}{n\choose{j}}\frac{d^{j}v}{dx^{j}}\frac{d^{n-j}u}{dx^{n-j}},\quad\forall{v,u}\in{C^{n}}(\mathbb{R}^{1}).$
(8)

This is the Leibniz rule. Similarly we have

$v\frac{d^{n}u}{dx^{n}}=\sum^{n}_{j=0}(-1)^{j}{n\choose{j}}\frac{d^{n-j}}{dx^{n-j}}\left(u\frac{d^{j}v}{dx^{j}}\right),\quad\forall{v,u}\in{C^{n}}(\mathbb{R}^{1}).$
(9)

$\frac{d^{k}}{dx^{k}}f(x^{2})=\sum^{[k/2]}_{j=0}\frac{k!}{j!\,(k-2j)!}(2x)^{k-2j}\frac{d^{k-j}}{dy^{k-j}}f(y),\quad\forall{f(x^{2})}\in{C^{k}}(\mathbb{R}^{1}),\;y=x^{2}.$
(10)

By (10) we easily obtain the Hermite polynomials

$H_{k}(x)=(-1)^{k}e^{x^{2}}\frac{d^{k}}{dx^{k}}e^{-x^{2}}=\sum^{[k/2]}_{j=0}(-1)^{j}\frac{k!}{j!\,(k-2j)!}(2x)^{k-2j}.$

The major parts of rules of differentiation are the derivative principle of
function product, the derivative principle of compound function and the chain
rule of multivariate function. Firstly let us establish the derivative
principle of function product. Without losing the universality, we only
consider the situation of the product of two functions. Let $n=1$ in (8), then
for $v=f(x)$ and $u=g(x)$ we have

$\frac{d}{dx}(f(x)g(x))=f(x)\frac{d}{dx}g(x)+g(x)\frac{d}{dx}f(x),\quad\forall{f,g}\in{C}^{1}(\mathbb{R}^{1}).$
(11)

Let $\varphi(x)=f_{1}(x)f_{2}(x)\cdots{f}_{n}(x)$, generally we have

$\frac{d}{dx}\varphi(x)=\sum^{n}_{j=1}\prod^{n}_{i=1\atop
i\neq{j}}f_{i}(x)\frac{d}{dx}f_{j}(x),\quad\forall{f_{j}(x)}\in{C}^{1}(\mathbb{R}^{1}).$

Let $f_{1}(x)=f_{2}(x)=\cdots=f_{n}(x)=g(x)=y$, then for differentiable
function $y=g(x)$ we have

$\frac{d}{dx}y^{n}=ny^{n-1}\frac{dy}{dx}\quad\mbox{or}\quad\frac{d}{dx}y^{n}=\frac{dy}{dx}\frac{d}{dy}y^{n},\qquad\forall{n}\in\mathbb{N}^{1}.$

According to the analytic continuous fundamental theorem, we have the
derivative principle of compound function $f(g(x))\in{C^{1}}(\mathbb{R}^{1})$:

$\frac{d}{dx}f(g(x))=\frac{dy}{dx}\frac{d}{dy}f(y),\qquad{y}=g(x),\quad{x}\in\mathbb{R}^{1}.$
(12)

By using (11) and (12), for differentiable functions $x_{1}(t)$ and
$x_{2}(t)$, $t\in\mathbb{R}^{1}$ we have

$\frac{d}{dt}(x^{\alpha_{1}}_{1}x^{\alpha_{2}}_{2})=\frac{dx^{\alpha_{1}}_{1}}{dx_{1}}\frac{dx_{1}}{dt}x^{\alpha_{2}}_{2}+x^{\alpha_{1}}_{1}\frac{dx^{\alpha_{2}}_{2}}{dx_{2}}\frac{dx_{2}}{dt},\qquad\alpha_{1},\alpha_{2}\in\mathbb{N}.$

Taking this one as the characteristic equation, by base functions
$x^{\alpha}$, $x\in\mathbb{R}^{2},\;\alpha\in\mathbb{N}^{2}$, combining it
with the following operator equation:

$\frac{d}{dt}x^{\alpha}=\frac{dx_{1}}{dt}\frac{\partial}{\partial{x}_{1}}x^{\alpha}+\frac{dx_{2}}{dt}\frac{\partial}{\partial{x}_{2}}x^{\alpha},\qquad{x}(t)\in\mathbb{R}^{2}\quad{t}\in\mathbb{R}^{1}\quad\alpha\in\mathbb{N}^{2}.$

According to the analytic continuous fundamental theorem, it is tenable
$\forall{f(x)}\in{C^{1}}(\mathbb{R}^{2})$, namely

$\frac{d}{dt}f(x)=\frac{dx_{1}}{dt}\frac{\partial}{\partial{x}_{1}}f(x)+\frac{dx_{2}}{dt}\frac{\partial}{\partial{x}_{2}}f(x),\qquad{x}(t)\in\mathbb{R}^{2}\quad{t}\in\mathbb{R}^{1}.$

Clearly, for $x(t)\in\mathbb{R}^{n}$, $t\in\mathbb{R}^{1}$, we can generally
derive $\forall{f(x)}\in{C^{1}}(\mathbb{R}^{n})$

$\frac{d}{dt}f(x)=\frac{dx_{1}}{dt}\frac{\partial}{\partial{x}_{1}}f(x)+\frac{dx_{2}}{dt}\frac{\partial}{\partial{x}_{2}}f(x)+\cdots+\frac{dx_{n}}{dt}\frac{\partial}{\partial{x}_{n}}f(x).$
(13)

If taking $n\in\mathbb{N}$ as the character of base function, the binomial
formula can be expressed as the following characteristic equation:

$(x+h)^{n}=\sum^{\infty}_{j=0}\frac{h^{j}}{j!}\frac{d^{j}}{dx^{j}}x^{n}.$

According to the analytic continuous fundamental theorem, it is tenable for
any analytic function $f(x)\in{C^{\infty}}(\mathbb{C}^{1})$, namely

$f(x+h)=\sum^{\infty}_{j=0}\frac{h^{j}}{j!}\frac{d^{j}}{dx^{j}}f(x),\quad{x}\in\mathbb{C}^{1}.$
(14)

This is the Taylor formula. If taking (14) as a characteristic equation, then
similarly we have

Theorem 1. If $\exists{v(s),u(s)}\in{C^{\infty}}(\mathbb{C}^{1})$ can make the
infinite series on the right side of (15) uniform convergent, then it will
uniform converges to the left side of the formula, namely

$f\\!\left(\lambda\frac{\partial}{\partial{s}}\right)(vu)=\sum_{k=0}^{\infty}\frac{\lambda^{k}}{k!}\,\frac{\partial^{k}v}{\partial{s^{k}}}\,f^{(k)}\\!\left(\lambda\frac{\partial}{\partial{s}}\right)u,\quad{s}\in\mathbb{C}^{1},\;|\lambda|\leq
1.$ (15)

Corollary 4. In (15), when $\lambda=-1,\;v=s,\;u=1/s,\,s\neq 0$, we have

$f^{\prime}\left(-\frac{\partial}{\partial{s}}\right)\frac{1}{s}=sf\left(-\frac{\partial}{\partial{s}}\right)\frac{1}{s}-f(0)\quad\mbox{or}\quad\mathscr{L}f^{\prime}(t)=s\mathscr{L}f(t)-f(0).$
(16)

Where $\mathscr{L}f(t)$ denotes

$\mathscr{L}f(t)=f\left(-\frac{\partial}{\partial{s}}\right)\frac{1}{s},\quad\;\Re(s)>0,\;t\in\overline{\mathbb{R}_{+}^{1}}.$
(17)

This is the internal connection between abstract operators and Laplace
transform, thus the abstract operator becomes a significant tool to compute
Laplace transform. For instance, by (17) we have

$\mathscr{L}[g(t)f(t)]=g\left(-\frac{\partial}{\partial{s}}\right)F(s),\quad(F(s)=\mathscr{L}f(t)).$
(18)

Similar to (11)-(13), Guangqing Bi has already established a series of
algorithms of abstract operators in reference [1]:

$\cos(h\partial_{x})f(x)=\Re[f(x+ih)],\quad\sin(h\partial_{x})f(x)=\Im[f(x+ih)],\quad\forall{f(z)}\in{C}^{\infty}(\mathbb{C}^{n}).$
(19)

Where
$bx=b_{1}x_{1}+b_{2}x_{2}+\cdots+b_{n}x_{n},\;bh=b_{1}h_{1}+b_{2}h_{2}+\cdots+b_{n}h_{n}$.

$\displaystyle\sin(h\partial_{x})(uv)$ $\displaystyle=$
$\displaystyle\cos(h\partial_{x})v\cdot\sin(h\partial_{x})u+\sin(h\partial_{x})v\cdot\cos(h\partial_{x})u,$
$\displaystyle\cos(h\partial_{x})(uv)$ $\displaystyle=$
$\displaystyle\cos(h\partial_{x})v\cdot\cos(h\partial_{x})u-\sin(h\partial_{x})v\cdot\sin(h\partial_{x})u.$

$\displaystyle\sin(h\partial_{x})\frac{u}{v}$ $\displaystyle=$
$\displaystyle\frac{\cos(h\partial_{x})v\cdot\sin(h\partial_{x})u-\sin(h\partial_{x})v\cdot\cos(h\partial_{x})u}{(\cos(h\partial_{x})v)^{2}+(\sin(h\partial_{x})v)^{2}},$
$\displaystyle\cos(h\partial_{x})\frac{u}{v}$ $\displaystyle=$
$\displaystyle\frac{\cos(h\partial_{x})v\cdot\cos(h\partial_{x})u+\sin(h\partial_{x})v\cdot\sin(h\partial_{x})u}{(\cos(h\partial_{x})v)^{2}+(\sin(h\partial_{x})v)^{2}}.$

Where $x\in\mathbb{R}^{n},\;h\in\mathbb{R}_{n}\,\mbox{or}\,\mathbb{C}_{n}$.

$\displaystyle\sin(h\partial_{t})f(x(t))$ $\displaystyle=$
$\displaystyle\sin(Y\partial_{X})f(X),$
$\displaystyle\cos(h\partial_{t})f(x(t))$ $\displaystyle=$
$\displaystyle\cos(Y\partial_{X})f(X).$

$\displaystyle
x(t)\in\mathbb{R}^{n},\;t\in\mathbb{R}^{1},\;h\in\mathbb{R}_{1}\,\mbox{or}\,\mathbb{C}_{1}.$

Where
$X=(X_{1},X_{2},\cdots,X_{n}),\;X_{j}=\cos(h\partial_{t})x_{j}(t),\;Y\in\mathbb{R}_{n},\;Y_{j}=\sin(h\partial_{t})x_{j}(t)$.

For example, when $n=1$, the (2.2) can easily be restated as

$\displaystyle\sin\left(h\frac{d}{dt}\right)f(x(t))$ $\displaystyle=$
$\displaystyle\sin\left(Y\frac{\partial}{\partial{X}}\right)f(X),$
$\displaystyle\cos\left(h\frac{d}{dt}\right)f(x(t))$ $\displaystyle=$
$\displaystyle\cos\left(Y\frac{\partial}{\partial{X}}\right)f(X).$

Where
$Y=\sin(h\partial_{t})x(t),\;X=\cos(h\partial_{t})x(t),\;t\in\mathbb{R}^{1},\;h\in\mathbb{R}_{1}\,\mbox{or}\,\mathbb{C}_{1}$.

If $n=2$, then (2.2) can easily be restated as

$\displaystyle\sin\left(h\frac{d}{dt}\right)f(x(t),y(t))$ $\displaystyle=$
$\displaystyle\sin\left(Y_{x}\frac{\partial}{\partial{X}_{x}}+Y_{y}\frac{\partial}{\partial{X}_{y}}\right)f(X_{x},X_{y}),$
$\displaystyle\cos\left(h\frac{d}{dt}\right)f(x(t),y(t))$ $\displaystyle=$
$\displaystyle\cos\left(Y_{x}\frac{\partial}{\partial{X}_{x}}+Y_{y}\frac{\partial}{\partial{X}_{y}}\right)f(X_{x},X_{y}).$

Where
$Y_{x}=\sin(h\partial_{t})x(t),\;X_{x}=\cos(h\partial_{t})x(t),\;Y_{y}=\sin(h\partial_{t})y(t),\;X_{y}=\cos(h\partial_{t})y(t)$.

Theorem 7 in reference [1] can be more generally expressed as:

Theorem 2. If $y=f(bx)$ in set of analytic functions is the inverse function
of $bx=g(y)$, namely $g(f(bx))=bx$, then $\sin(h\partial_{x})f(bx)$(denoted by
$Y$) and $\cos(h\partial_{x})f(bx)$(denoted by $X$) can be determined by the
following set of equations:

$\displaystyle\cos\left(Y\frac{\partial}{\partial{X}}\right)g(X)$
$\displaystyle=$ $\displaystyle bx,$
$\displaystyle\sin\left(Y\frac{\partial}{\partial{X}}\right)g(X)$
$\displaystyle=$ $\displaystyle bh.$

$\displaystyle
x\in\mathbb{R}^{n},\quad{b}\in\mathbb{R}_{n},\quad{h}\in\mathbb{R}_{n}.$

By using (2.2)-(2.2), we can constructively extend the domains and ranges of
the following five abstract operators

$\exp(h\partial_{x}),\;\sin(h\partial_{x}),\;\cos(h\partial_{x}),\;\sinh(h\partial_{x})\;\mbox{and}\;\cosh(h\partial_{x})$

from power functions and exponential functions to all the $C^{\infty}$
functions. For instance (See [5])

$\displaystyle\sin(h\partial_{x})\arctan{bx}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\textrm{tanh}^{-1}\frac{2bh}{1+(bx)^{2}+(bh)^{2}},$
$\displaystyle\cos(h\partial_{x})\arctan{bx}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\arctan\frac{2bx}{1-(bx)^{2}-(bh)^{2}}.$

Therefore, establishing such a set of algorithms as (2.2)-(2.2) is the second
important way to extend the domains and ranges of abstract operators.

Because of the introduction of Definition 5, it is necessary to add new
algorithms.

Theorem 3. Let $a\in\mathbb{C}^{1}$, $\rho\in\mathbb{R}^{1}$, then
$\forall{f}(z)\in{C}^{\infty}(\mathbb{C}^{1})$, we have

$a^{\rho\frac{\partial}{\partial\rho}}f(\rho)=f(a\rho).$ (27)

Proof. By Definition 5, for $a\in\mathbb{C}^{1}$, $\rho\in\mathbb{R}^{1}$, we
have

$a^{\rho\frac{\partial}{\partial\rho}}\rho^{\alpha}=a^{\alpha}\rho^{\alpha}\quad\mbox{or}\quad
a^{\rho\frac{\partial}{\partial\rho}}\rho^{\alpha}=(a\rho)^{\alpha},\qquad\forall{\alpha\in\mathbb{N}^{1}}.$

According to the analytic continuous fundamental theorem, we have Theorem 3.

Theorem 4. Let
$X=(\rho_{1}\cos\theta_{1},\cdots,\rho_{n}\cos\theta_{n}),\;Y=(\rho_{1}\sin\theta_{1},\cdots,\rho_{n}\sin\theta_{n})$,
introducing the notation

$\theta\,\rho\frac{\partial}{\partial\rho}=\theta_{1}\,\rho_{1}\frac{\partial}{\partial\rho_{1}}+\cdots+\theta_{n}\,\rho_{n}\frac{\partial}{\partial\rho_{n}},\;\;Y\partial_{X}=Y_{1}\frac{\partial}{\partial{X}_{1}}+\cdots+Y_{n}\frac{\partial}{\partial{X}_{n}},$

then
$\forall{f}(z)\in{C}^{\infty}(\mathbb{C}^{n}),\;\rho\in\mathbb{R}^{n},\;\theta\in\mathbb{R}_{n}$,
we have

$\displaystyle\cos\left(\theta\,\rho\frac{\partial}{\partial\rho}\right)f(\rho)$
$\displaystyle=$ $\displaystyle\cos(Y\partial_{X})f(X),$
$\displaystyle\sin\left(\theta\,\rho\frac{\partial}{\partial\rho}\right)f(\rho)$
$\displaystyle=$ $\displaystyle\sin(Y\partial_{X})f(X).$

Proof. Let $a_{1}=e^{i\theta_{1}},\,\ldots,\,a_{n}=e^{i\theta_{n}}$, according
to Theorem 3, then

$\exp\left(i\theta\,\rho\frac{\partial}{\partial\rho}\right)f(\rho)=f(\rho{e}^{i\theta})=f(X+iY)=\exp(iY\partial_{X})f(X),\qquad
X\in\mathbb{R}^{n},\;\;Y\in\mathbb{R}_{n}.$

Considering

$\displaystyle\exp\left(i\theta\,\rho\frac{\partial}{\partial\rho}\right)$
$\displaystyle=$
$\displaystyle\cos\left(\theta\,\rho\frac{\partial}{\partial\rho}\right)+i\sin\left(\theta\,\rho\frac{\partial}{\partial\rho}\right),$
$\displaystyle\exp(iY\partial_{X})$ $\displaystyle=$
$\displaystyle\cos(Y\partial_{X})+i\sin(Y\partial_{X}),$

thus we have Theorem 4, which transforms the abstract operators taking
$\rho\frac{\partial}{\partial\rho}$ as the operator element into those taking
$\partial_{X}$ as the operator element.

Corollary 5. Let $f(z),\,z=x+iy\in\mathbb{C}^{1}$ be an arbitrary analytic
function in a complex plane, then on the line $y=kx$ we have

$\displaystyle\left.\cos\left(y\frac{\partial}{\partial{x}}\right)f(x)\right|_{y=kx}$
$\displaystyle=$
$\displaystyle\beta^{x\frac{\partial}{\partial{x}}}\cos\left(\alpha{x}\frac{\partial}{\partial{x}}\right)f(x),$
$\displaystyle\left.\sin\left(y\frac{\partial}{\partial{x}}\right)f(x)\right|_{y=kx}$
$\displaystyle=$
$\displaystyle\beta^{x\frac{\partial}{\partial{x}}}\sin\left(\alpha{x}\frac{\partial}{\partial{x}}\right)f(x).$

Where $\beta=(1+k^{2})^{1/2},\;\alpha=\arctan{k}$, which can be used to solve
boundary value problems of 2-dimensional Laplace equation on polygonal
domains.

Corollary 6. Making use of the analytic continuous fundamental theorem and the
Pochhammer symbol $(\lambda)_{m}$ defined by

$(\lambda)_{m}=\frac{\Gamma(\lambda+m)}{\Gamma(\lambda)}=\left\\{\begin{array}[]{l@{\qquad}l}\displaystyle
1,&(m=0),\\\
\displaystyle\lambda(\lambda+1)\cdots(\lambda+m-1)&(m\in\mathbb{N}_{+}).\end{array}\right.$
(30)

It is easily seen

$(-X)_{m}u(x)=(-1)^{m}x^{m}\frac{d^{m}u}{dx^{m}},\quad\forall{u}(x)\in{C^{m}}(\mathbb{R}^{1}),\,X=x\frac{d}{dx}=\frac{d}{dz}=D_{z}.$
(31)

Where $x=e^{z}$ or $z=\ln{x}$. Using this set of formulas, we can easily
transform the Euler equation into an ordinary differential equation with
constant coefficients.

Therefore, for an arbitrary abstract operators, its symbol are $C^{\infty}$
functions. Further descriptions require concrete analysis under specific
circumstances. For instance, the symbol of the abstract operators on the right
side of (17) can be further described by using conditions of the Laplace
transform. The domain of the abstract operators $P(\partial_{x})$,
$\sin(h\partial_{x})\;\mbox{and}\;\cos(h\partial_{x})$ are the $C^{\infty}$
whole space.

###  2.3 Abstract operators and bounded function

The abstract operators can also acting on the differentiable functions on
bounded domain.

For instance, let $a\leq{b}$, without losing the universality, assuming $a\geq
0,\;b\geq 0$ we have

$na^{n-1}\leq(a^{n-1}+a^{n-2}b+a^{n-3}b^{2}+\cdots+ab^{n-2}+b^{n-1})\leq{n}b^{n-1}.$
$na^{n-1}\leq\frac{b^{n}-a^{n}}{b-a}\leq{n}b^{n-1}\quad\mbox{or}\quad{a}\leq\left(\frac{1}{n}\frac{b^{n}-a^{n}}{b-a}\right)^{\frac{1}{n-1}}\leq{b}.$

Therefore, if $a\leq{b}$, then we have $a\leq{c}\leq{b}$, making

$nc^{n-1}=\frac{b^{n}-a^{n}}{b-a}\quad\mbox{or}\quad\frac{d}{dc}c^{n}=\frac{b^{n}-a^{n}}{b-a},\qquad\forall{n}\in\mathbb{N}^{1}.$

Taking this one as the characteristic equation, according to the analytic
continuous fundamental theorem we get the Lagrange mean value theorem, namely

If $a\leq{b}$, and $f(x)\in{C^{1}}[a,b]$, then $\exists{c}\in[a,b]$ making

$\frac{d}{dc}f(c)=\frac{f(b)-f(a)}{b-a},\quad\forall{f(x)}\in{C^{1}}[a,b].$
(32)

In reference [5], Guangqing Bi and Yuekai Bi discussed the following
functions:

$\displaystyle f(x)$ $\displaystyle=$
$\displaystyle\left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)S(e^{z})\right|_{z=0},$
$\displaystyle g(x)$ $\displaystyle=$
$\displaystyle\left.\sin\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)S(e^{z})\right|_{z=0},\quad{z}\in\mathbb{R}^{1},\;a<x<b.$

Where $f(x),g(x)\in{C}_{0}^{\infty}(\mathbb{R}^{1})$,
$S(t)\in{C}^{\infty}(\Omega_{0})$, the $\Omega_{0}$ is a certain neighborhood
at $t=0$.

If $S(t)$ is expanded in power series on $\Omega_{0}$, then (2.3) are
equivalent to Fourier series, and their domain $(a,b)$ can be extended to all
$(-\infty,+\infty)$ periodically. So $f(x),g(x)\in{L}^{2}([-c,c])$ and
$f(x+2l)=f(x),\;g(x+2l)=g(x)$.

For instance, by (2.2) and (2.2), we have

$\displaystyle\left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)\arctan{e^{z}}\right|_{z=0}\,=\,\left.\cos\left(Y\frac{\partial}{\partial{X}}\right)\arctan{X}\right|_{z=0}$
$\displaystyle=$
$\displaystyle\left.\frac{1}{2}\arctan\frac{2X}{1-(X^{2}+Y^{2})}\right|_{z=0}\,=\,\frac{1}{2}\arctan\frac{2\cos(\pi{x}/c)}{1-\left(\cos^{2}(\pi{x}/c)+\sin^{2}(\pi{x}/c)\right)}.$

When $\cos(\pi{x}/c)\neq 0$ or $|x|\neq{kc}+c/2,\;k=0,1,2,\cdots$, the above
expression turns into

$\left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)\arctan{e^{z}}\right|_{z=0}=\left\\{\begin{array}[]{r@{\qquad}l}(1/2)\arctan(+\infty)=+\pi/4,&2Kc-c/2<x<2Kc+c/2\\\
(1/2)\arctan(-\infty)=-\pi/4,&2Kc+c/2<x<2Kc+3c/2.\end{array}\right.$

Where $K=0,\pm 1,\pm 2,\cdots$. This function represents a square wave.

By algorithms (19)-(2.2), and some basic formulas similar to (2.2) (See [5]),
we obtained

$\displaystyle\left.\sin\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)(-\ln(1-e^{z}))\right|_{z=0}$
$\displaystyle=$ $\displaystyle\frac{\pi}{2}-\frac{\pi{x}}{2c},\quad 0<x<2c.$
$\displaystyle\left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)(-\ln(1-e^{z}))\right|_{z=0}$
$\displaystyle=$ $\displaystyle-\ln\left(2\sin\frac{\pi{x}}{2c}\right),\quad
0<x<2c.$
$\displaystyle\left.\sin\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)\ln(1+e^{z})\right|_{z=0}$
$\displaystyle=$ $\displaystyle\frac{\pi{x}}{2c},\quad|x|<c.$
$\displaystyle\left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)\ln(1+e^{z})\right|_{z=0}$
$\displaystyle=$ $\displaystyle\ln\left(2\cos\frac{\pi
x}{2c}\right),\quad|x|<c.$
$\displaystyle\left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)\arctan{e^{z}}\right|_{z=0}$
$\displaystyle=$ $\displaystyle\frac{\pi}{4},\quad|x|<c/2.$

In other words, the ranges of the abstract operators $\sin(h\partial_{x})$ and
$\cos(h\partial_{x})$ may be extended from the
$C_{0}^{\infty}(\mathbb{R}^{1})$ to the $L^{2}([-c,c])$.

In order to solve an initial-boundary value problems of partial differential
equations, we need to express a function with boundary conditions. For
instance, (2.3) are infinitely differentiable within a finite domain, such as

$\frac{d^{k}}{dx^{k}}\left.\sin\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)(-\ln(1-e^{z}))\right|_{z=0}=\frac{d^{k}}{dx^{k}}\left(\frac{\pi}{2}-\frac{\pi{x}}{2c}\right),\quad\forall{k}\in\mathbb{N},\;0<x<2c.$

Example 3. $\forall{f(z)}\in{C}^{\infty}(\mathbb{C}^{1})$, if $f(z)$ is
continuous at $z=1/2$, then

$\frac{2}{\pi}\int^{\pi/2}_{0}\\!\left.\cos\left(x\,\rho\frac{\partial}{\partial\rho}\right)f(\rho)\right|_{\rho=\cos{x}}dx=f\left(\frac{1}{2}\right).$
(35)

Proof. Taking the following integral formula as the characteristic equation

$\int^{\pi/2}_{0}\\!\cos^{n}x\,\cos{nx}\,dx=\frac{\pi}{2^{n+1}},\quad\forall{n}\in\mathbb{N}.$

Combining the characteristic equation with the following operator equation by
using Definition 5 and the base function $x^{\alpha}$:

$\frac{2}{\pi}\int^{\pi/2}_{0}\\!\cos\left(x\,\rho\frac{\partial}{\partial\rho}\right)\rho^{n}\,dx=\left(\frac{1}{2}\right)^{n},\qquad\forall{n}\in\mathbb{N}^{1},\quad
x\in[0,\pi/2].$

Where $\rho=\cos{x}$, according to the analytic continuous fundamental
theorem, we get (35).

Example 4. $\forall{f(z)}\in{C}^{\infty}(\mathbb{C}^{1})$, if $f(z)$ is
differentiable at $z=1/2$, then

$\int^{\pi/2}_{0}\\!\left.\sin\left(x\,\rho\frac{\partial}{\partial\rho}\right)f(\rho)\right|_{\rho=\cos{x}}dx=\frac{1}{2}\int^{1}_{0}\\!\frac{f(\xi)-f(1/2)}{\xi-1/2}\,d\xi.$
(36)

Proof. Considering another integral formula

$\int^{\pi/2}_{0}\\!\cos^{n}x\,\sin{nx}\,dx=\frac{1}{2^{n+1}}\sum^{n}_{k=1}\frac{2^{k}}{k},\quad\forall{n}\in\mathbb{N}.$

Obviously it cannot be taken as the characteristic equation directly, but
considering $\sum^{n}_{k=1}2^{k}/k=\int^{2}_{0}\frac{q^{n}-1}{q-1}dq$, and
according to the analytic continuous fundamental theorem, we get (36).

Example 5. Let $k\in\mathbb{N}_{+}$,
$\forall{f(z)}\in{C}^{\infty}(\mathbb{C}^{1})$, if $f(z)$ is $k$ times
continuously differentiable at $z=1/2$, then

$\frac{2}{\pi}\int^{\pi/2}_{0}\\!\cos(2kx)\left.\cos\left(x\,\rho\frac{\partial}{\partial\rho}\right)f(\rho)\right|_{\rho=\cos{x}}dx=\frac{1}{2^{k+1}k!}f^{(k)}\left(\frac{1}{2}\right).$
(37)
$\frac{2}{\pi}\int^{\pi/2}_{0}\\!\sin(2kx)\left.\sin\left(x\,\rho\frac{\partial}{\partial\rho}\right)f(\rho)\right|_{\rho=\cos{x}}dx=\frac{1}{2^{k+1}k!}f^{(k)}\left(\frac{1}{2}\right).$
(38)

Proof. In [6], Connon has given the following useful identical
equations$(\forall{n}\in\mathbb{N})$:

$2^{n}\cos^{n}x\cos(nx)=\sum^{n}_{k=0}{n\choose{k}}\cos(2kx).$ (39)
$2^{n}\cos^{n}x\sin(nx)=\sum^{n}_{k=0}{n\choose{k}}\sin(2kx).$ (40)

Using the orthogonality of trigonometric functions, by (39) and (40) we have

$\frac{1}{\pi}\int_{0}^{\pi}\cos(2kx)\cos^{n}x\cos(nx)dx={n\choose{k}}\left(\frac{1}{2}\right)^{n+1},\quad\forall{k}\in\mathbb{N}_{+}.$
(41)
$\frac{1}{\pi}\int_{0}^{\pi}\sin(2kx)\cos^{n}x\sin(nx)dx={n\choose{k}}\left(\frac{1}{2}\right)^{n+1},\quad\forall{k}\in\mathbb{N}_{+}.$
(42)

Which can also be written as

$\frac{2}{\pi}\int_{0}^{\pi/2}\cos(2kx)\cos^{n}x\cos(nx)dx={n\choose{k}}\left(\frac{1}{2}\right)^{n+1}.$
(43)
$\frac{2}{\pi}\int_{0}^{\pi/2}\sin(2kx)\cos^{n}x\sin(nx)dx={n\choose{k}}\left(\frac{1}{2}\right)^{n+1}.$
(44)

Clearly, (43) is the characteristic equation of (37), and 44) is the
characteristic equation of (38). So by Definition 5 we have

$\frac{2}{\pi}\int^{\pi/2}_{0}\\!\cos(2kx)\left.\cos\left(x\,\rho\frac{\partial}{\partial\rho}\right)\rho^{n}\right|_{\rho=\cos{x}}dx=\frac{1}{2^{k+1}k!}\left.\frac{d^{k}}{d\xi^{k}}\xi^{n}\right|_{\xi=1/2},\quad\forall{n}\in\mathbb{N}^{1}.$
(45)
$\frac{2}{\pi}\int^{\pi/2}_{0}\\!\sin(2kx)\left.\sin\left(x\,\rho\frac{\partial}{\partial\rho}\right)\rho^{n}\right|_{\rho=\cos{x}}dx=\frac{1}{2^{k+1}k!}\left.\frac{d^{k}}{d\xi^{k}}\xi^{n}\right|_{\xi=1/2},\quad\forall{n}\in\mathbb{N}^{1}.$
(46)

Then according to the analytic continuous fundamental theorem, we get (37) and
(38).

In reference [5] we get

Let
$S(t)=\sum^{\infty}_{n=0}a_{n}t^{n},\;t\in\mathbb{R}^{1},\;0\leq{t}\leq{r},\;0<r<+\infty$,
if

$\sum^{\infty}_{n=0}a_{n}\cos\frac{n\pi{x}}{c}=\left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)S(e^{z})\right|_{z=0},$
$\sum^{\infty}_{n=0}a_{n}\sin\frac{n\pi{x}}{c}=\left.\sin\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)S(e^{z})\right|_{z=0}$

are tenable in $a<x<b,\;x\in\mathbb{R}^{1}$, then
$\forall{x_{0}}\in[0,\frac{b-a}{2})$,

$\displaystyle\sum^{\infty}_{n=0}a_{n}\cos\frac{n\pi{x_{0}}}{c}\cos\frac{n\pi{x}}{c}$
$\displaystyle=$
$\displaystyle\cosh\left(x_{0}\frac{\partial}{\partial{x}}\right)\left.\cos\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)S(e^{z})\right|_{z=0},$
$\displaystyle\sum^{\infty}_{n=0}a_{n}\cos\frac{n\pi{x_{0}}}{c}\sin\frac{n\pi{x}}{c}$
$\displaystyle=$
$\displaystyle\cosh\left(x_{0}\frac{\partial}{\partial{x}}\right)\left.\sin\left(\frac{\pi{x}}{c}\frac{\partial}{\partial{z}}\right)S(e^{z})\right|_{z=0}$

are tenable in $a+x_{0}<x<b-x_{0}$. Accordingly, for any definite value
$x\in(a,b)$, $x_{0}$ in (2.3) takes values in the following interval:

$0<x_{0}<\left\\{\begin{array}[]{lr}x-a,&a<x\leq(a+b)/2,\\\
b-x,&b>x\geq(a+b)/2.\end{array}\right.$

Where the left side of (2.3) is:

$\cos\frac{n\pi{x_{0}}}{c}\cos\frac{n\pi{x}}{c}=\cosh\left(x_{0}\frac{\partial}{\partial{x}}\right)\cos\frac{n\pi{x}}{c},$
$\cos\frac{n\pi{x_{0}}}{c}\sin\frac{n\pi{x}}{c}=\cosh\left(x_{0}\frac{\partial}{\partial{x}}\right)\sin\frac{n\pi{x}}{c}.$

Therefore, $\sinh(h\partial_{x})\;\mbox{and}\;\cosh(h\partial_{x})$ are the
continuous operators on $L^{2}([-c,c])$.

###  2.4 Abstract operators and Fourier analysis

The third way to extend the domains and ranges of abstract operators is using
Fourier analysis.

Theorem 5. Let $t\in\overline{\mathbb{R}_{+}^{1}}$,
$\forall{f(x)}\in{S}(\mathbb{R}^{1})$, if $F_{+}(s),\,F_{-}(s)$ are
respectively

$F_{+}(s)=\frac{1}{\pi}\mathscr{L}\int^{\infty}_{-\infty}f(\xi)\cos(t\xi)d\xi,\qquad
F_{-}(s)=\frac{1}{\pi}\mathscr{L}\int^{\infty}_{-\infty}f(\xi)\sin(t\xi)d\xi,$
(48)

where $\mathscr{L}$ is the Laplace transform operator, then we have

$f(x)=\left.\cos\left(x\frac{\partial}{\partial{s}}\right)F_{+}(s)\right|_{s=0}+\left.\sin\left(-x\frac{\partial}{\partial{s}}\right)F_{-}(s)\right|_{s=0}.$
(49)

Or

$\left.\cos\left(x\frac{\partial}{\partial{s}}\right)F_{+}(s)\right|_{s=0}=\frac{f(x)+f(-x)}{2},\qquad\frac{1}{\pi}\int^{\infty}_{-\infty}f(x)\cos(tx)dx=\mathscr{L}^{-1}F_{+}(s).$
(50)
$\left.\sin\left(-x\frac{\partial}{\partial{s}}\right)F_{-}(s)\right|_{s=0}=\frac{f(x)-f(-x)}{2},\qquad\frac{1}{\pi}\int^{\infty}_{-\infty}f(x)\sin(tx)dx=\mathscr{L}^{-1}F_{-}(s).$
(51)

Proof. By (17) we have

$\mathscr{L}\cos(t\xi)=\cos\left(\xi\frac{\partial}{\partial{s}}\right)\frac{1}{s},\quad\mathscr{L}\sin(t\xi)=\sin\left(-\xi\frac{\partial}{\partial{s}}\right)\frac{1}{s},\;\;t\in\overline{\mathbb{R}_{+}^{1}},\;\Re(s)>0.$

So by (48) and (49),
$\forall{f(x)}\in{S}(\mathbb{R}^{1}),\,\exists{f_{s}(x)}\in{C^{\infty}}(\mathbb{R}^{1})$
can be expressed as

$\displaystyle f_{s}(x)$ $\displaystyle=$
$\displaystyle\cos\left(x\frac{\partial}{\partial{s}}\right)F_{+}(s)+\sin\left(-x\frac{\partial}{\partial{s}}\right)F_{-}(s)$
$\displaystyle=$
$\displaystyle\cos\left(x\frac{\partial}{\partial{s}}\right)\left[\frac{1}{\pi}\int^{\infty}_{-\infty}f(\xi)\mathscr{L}\cos(t\xi)d\xi\right]+\sin\left(-x\frac{\partial}{\partial{s}}\right)\left[\frac{1}{\pi}\int^{\infty}_{-\infty}f(\xi)\mathscr{L}\sin(t\xi)d\xi\right]$
$\displaystyle=$
$\displaystyle\frac{1}{\pi}\cos\left(x\frac{\partial}{\partial{s}}\right)\int^{\infty}_{-\infty}f(\xi)\cos\left(\xi\frac{\partial}{\partial{s}}\right)\frac{1}{s}\,d\xi+\frac{1}{\pi}\sin\left(-x\frac{\partial}{\partial{s}}\right)\int^{\infty}_{-\infty}f(\xi)\sin\left(-\xi\frac{\partial}{\partial{s}}\right)\frac{1}{s}\,d\xi$
$\displaystyle=$
$\displaystyle\frac{1}{\pi}\int^{\infty}_{-\infty}f(\xi)\left[\cos\left(x\frac{\partial}{\partial{s}}\right)\cos\left(\xi\frac{\partial}{\partial{s}}\right)+\sin\left(-x\frac{\partial}{\partial{s}}\right)\sin\left(-\xi\frac{\partial}{\partial{s}}\right)\right]\frac{1}{s}\,d\xi$
$\displaystyle=$
$\displaystyle\frac{1}{\pi}\int^{\infty}_{-\infty}f(\xi)\cos\left((x-\xi)\frac{\partial}{\partial{s}}\right)\frac{1}{s}\,d\xi=\frac{1}{\pi}\int^{\infty}_{-\infty}f(\xi)\frac{s}{(x-\xi)^{2}+s^{2}}d\xi,\;\Re(s)>0.$

Therefore, we get

$\displaystyle\lim_{s\rightarrow 0^{+}}f_{s}(x)$ $\displaystyle=$
$\displaystyle\lim_{s\rightarrow
0^{+}}\frac{1}{\pi}\int^{\infty}_{-\infty}f(\xi)\frac{s}{(x-\xi)^{2}+s^{2}}d\xi$
$\displaystyle=$ $\displaystyle\int^{\infty}_{-\infty}f(\xi)\lim_{s\rightarrow
0^{+}}\frac{1}{\pi}\frac{s}{(x-\xi)^{2}+s^{2}}d\xi\rightharpoonup\int^{\infty}_{-\infty}f(\xi)\delta(x-\xi)d\xi=f(x).$

In the same manner we find

Theorem 6. Let ${f(x)}\in{L}^{2}([-l,l]),\;f(x+2l)=f(x)$. If $S_{+}(t)$ and
$S_{-}(t)$ are respectively

$S_{+}(t)=\frac{1}{2l}\int^{l}_{-l}f(\xi)\frac{1-t^{2}}{1-2t\cos(\pi\xi/l)+t^{2}}d\xi,$
(52)
$S_{-}(t)=\frac{1}{l}\int^{l}_{-l}f(\xi)\frac{t\sin(\pi\xi/l)}{1-2t\cos(\pi\xi/l)+t^{2}}d\xi,$
(53)

then $\exists{f_{z}(x)}\in{C^{\infty}}(\mathbb{R}^{1})$ with the following
form

$f_{z}(x)=\cos\left(\frac{\pi{x}}{l}\frac{\partial}{\partial{z}}\right)S_{+}(e^{z})+\sin\left(\frac{\pi{x}}{l}\frac{\partial}{\partial{z}}\right)S_{-}(e^{z}),\quad-\infty<z<0$
(54)

making

$\lim_{z\rightarrow
0^{-}}f_{z}(x)\rightharpoonup{f(x)},\quad\forall{f(x)}\in{C}(-l,l)\subset{L}^{2}([-l,l]).$
$\lim_{z\rightarrow
0^{-}}\|f(x)-f_{z}(x)\|_{L^{2}([-l,l])}=0,\quad\forall{f(x)}\in{L}^{2}([-l,l]).$

Proof. By the algorithms (2.2) we have

$\cos\left(\frac{\pi\xi}{l}\frac{\partial}{\partial{z}}\right)\frac{1+e^{z}}{1-e^{z}}=\frac{1-e^{2z}}{1-2e^{z}\cos(\pi\xi/l)+e^{2z}}.$
$\sin\left(\frac{\pi\xi}{l}\frac{\partial}{\partial{z}}\right)\frac{e^{z}}{1-e^{z}}=\frac{e^{z}\sin(\pi\xi/l)}{1-2e^{z}\cos(\pi\xi/l)+e^{2z}}.$

So $f_{z}(x)$ can be expressed as

$\displaystyle f_{z}(x)$ $\displaystyle=$
$\displaystyle\cos\left(\frac{\pi{x}}{l}\frac{\partial}{\partial{z}}\right)\left[\frac{1}{2l}\int^{l}_{-l}f(\xi)\cos\left(\frac{\pi\xi}{l}\frac{\partial}{\partial{z}}\right)\frac{1+e^{z}}{1-e^{z}}d\xi\right]$
$\displaystyle+\,\sin\left(\frac{\pi{x}}{l}\frac{\partial}{\partial{z}}\right)\left[\frac{1}{l}\int^{l}_{-l}f(\xi)\sin\left(\frac{\pi\xi}{l}\frac{\partial}{\partial{z}}\right)\frac{e^{z}}{1-e^{z}}d\xi\right]$
$\displaystyle=$
$\displaystyle\frac{1}{2l}\int^{l}_{-l}f(\xi)\cos\left(\frac{\pi{x}}{l}\frac{\partial}{\partial{z}}\right)\cos\left(\frac{\pi\xi}{l}\frac{\partial}{\partial{z}}\right)\left[1+\frac{2e^{z}}{1-e^{z}}\right]d\xi$
$\displaystyle+\,\frac{1}{l}\int^{l}_{-l}f(\xi)\sin\left(\frac{\pi{x}}{l}\frac{\partial}{\partial{z}}\right)\sin\left(\frac{\pi\xi}{l}\frac{\partial}{\partial{z}}\right)\frac{e^{z}}{1-e^{z}}d\xi=\frac{1}{2l}\int^{l}_{-l}f(\xi)d\xi$
$\displaystyle+\,\frac{1}{l}\int^{l}_{-l}f(\xi)\left[\cos\left(\frac{\pi{x}}{l}\frac{\partial}{\partial{z}}\right)\cos\left(\frac{\pi\xi}{l}\frac{\partial}{\partial{z}}\right)+\sin\left(\frac{\pi{x}}{l}\frac{\partial}{\partial{z}}\right)\sin\left(\frac{\pi\xi}{l}\frac{\partial}{\partial{z}}\right)\right]\frac{e^{z}}{1-e^{z}}d\xi$
$\displaystyle=$
$\displaystyle\frac{1}{2l}\int^{l}_{-l}f(\xi)d\xi+\frac{1}{l}\int^{l}_{-l}f(\xi)\cos\left(\frac{\pi(x-\xi)}{l}\frac{\partial}{\partial{z}}\right)\frac{e^{z}}{1-e^{z}}d\xi$
$\displaystyle=$
$\displaystyle\frac{1}{2l}\int^{l}_{-l}f(\xi)\cos\left(\frac{\pi(x-\xi)}{l}\frac{\partial}{\partial{z}}\right)\frac{1+e^{z}}{1-e^{z}}d\xi$
$\displaystyle=$
$\displaystyle\frac{1}{2l}\int^{l}_{-l}f(\xi)\frac{1-e^{2z}}{1-2e^{z}\cos(\pi(x-\xi)/l)+e^{2z}}d\xi,\quad-\infty<z<0.$

Therefore, if $f(x)\in{C}(-l,l)\subset{L}^{2}([-l,l])$, we get

$\displaystyle\lim_{z\rightarrow 0^{-}}f_{z}(x)$ $\displaystyle=$
$\displaystyle\lim_{z\rightarrow
0^{-}}\frac{1}{2l}\int^{l}_{-l}f(\xi)\frac{1-e^{2z}}{1-2e^{z}\cos(\pi(x-\xi)/l)+e^{2z}}d\xi$
$\displaystyle=$ $\displaystyle\int^{l}_{-l}f(\xi)\lim_{z\rightarrow
0^{-}}\frac{1}{2l}\frac{1-e^{2z}}{1-2e^{z}\cos(\pi(x-\xi)/l)+e^{2z}}d\xi\rightharpoonup\int^{l}_{-l}f(\xi)\delta(x-\xi)d\xi=f(x).$

As we all know, $\forall{f(x)}\in{L}^{2}([-l,l]),\;f(x+2l)=f(x)$ we have

$f(x)=a_{0}+\sum^{\infty}_{k=1}\left(a_{k}\cos\frac{k\pi{x}}{l}+b_{k}\sin\frac{k\pi{x}}{l}\right).$

Where $\forall{k}\in\mathbb{N}_{+}$

$a_{0}=\frac{1}{2l}\int^{l}_{-l}f(\xi)d\xi,\quad{a_{k}=\frac{1}{l}\int^{l}_{-l}f(\xi)\cos\frac{k\pi{x}}{l}d\xi},\quad{b_{k}=\frac{1}{l}\int^{l}_{-l}f(\xi)\sin\frac{k\pi{x}}{l}d\xi}.$

Accordingly $f_{z}(x)\in{C}^{\infty}(\mathbb{R}^{1})$ can be expressed as

$f_{z}(x)=a_{0}+\sum^{\infty}_{k=1}\left(a_{k}e^{kz}\cos\frac{k\pi{x}}{l}+b_{k}e^{kz}\sin\frac{k\pi{x}}{l}\right),\quad-\infty<z<0.$
(55)

So we have

$\frac{1}{l}\int^{l}_{-l}|f(x)-f_{z}(x)|^{2}dx=\sum^{\infty}_{k=1}[(a^{2}_{k}+b^{2}_{k})(1-e^{kz})^{2}],\quad-\infty<z<0.$
(56) $\lim_{z\rightarrow
0^{-}}\|f(x)-f_{z}(x)\|_{L^{2}([-l,l])}=0,\quad\forall{f(x)}\in{L}^{2}([-l,l]),\;f(x+2l)=f(x).$
(57)

If $f(x)$ is an even function, then we have the integral equation:

$\frac{1}{l}\int^{l}_{0}f(x)\frac{1-t^{2}}{1-2t\cos(\pi{x}/l)+t^{2}}dx=S_{+}(t),\quad\forall{S_{+}(t)}\in{C}^{\infty}(\Omega_{0}).$
(58)

Its solution $f(x)\in{L}^{2}([-l,l]),\;f(x+2l)=f(x)$, and we have

$f(x)=\left.\cos\left(\frac{\pi{x}}{l}\frac{\partial}{\partial{z}}\right)S_{+}(e^{z})\right|_{z=0}.$
(59)

If $f(x)$ is an odd function, then we have the integral equation:

$\frac{2}{l}\int^{l}_{0}f(x)\frac{t\sin(\pi{x}/l)}{1-2t\cos(\pi{x}/l)+t^{2}}dx=S_{-}(t),\quad\forall{S_{-}(t)}\in{C}^{\infty}(\Omega_{0}).$
(60)

Its solution $f(x)\in{L}^{2}([-l,l]),\;f(x+2l)=f(x)$, and we have

$f(x)=\left.\sin\left(\frac{\pi{x}}{l}\frac{\partial}{\partial{z}}\right)S_{-}(e^{z})\right|_{z=0}.$
(61)

For instance, in the integral equation (58), if $S_{+}(t)=\arctan{t},\;0<t<1$,
then its solution is a square wave, namely

$f(x)=\left\\{\begin{array}[]{r@{\qquad}l}+\pi/4,&2Kl-l/2<x<2Kl+l/2\\\
-\pi/4,&2Kl+l/2<x<2Kl+3l/2.\end{array}\right.$

Where $K=0,\pm 1,\pm 2,\cdots$.

Let $\rho_{z}(x)\in{C}_{0}^{\infty}(\mathbb{R}^{n})$,
supp$\rho_{z}(x)\subset\\{x\in\mathbb{R}^{n}|\,|x|\leq{l}\\}$ be a polished
kernel, taken as
$\rho_{z}(x)=\rho_{z_{1}}(x_{1})\rho_{z_{2}}(x_{2})\cdots\rho_{z_{i}}(x_{i})\cdots\rho_{z_{n}}(x_{n})$
and

$\rho_{z_{i}}(x_{i})=\frac{1}{2l_{i}}\frac{1-e^{2z_{i}}}{1-2e^{z_{i}}\cos(\pi{x_{i}}/l_{i})+e^{2{z_{i}}}},\quad-\infty<z_{i}<0,\;|x_{i}|\leq{l_{i}}\;(i=1,2,\cdots,n).$
(62)

Then we easily prove the following results:

$\lim_{z\rightarrow
0^{-}}g(\partial_{x})\int^{l}_{-l}f(\xi)\rho_{z}(x-\xi)d\xi\rightharpoonup{g}(\partial_{x})f(x),\quad\forall{f(x)}\in{L}^{2}([-l,l]),\;g(\xi)\in{C}^{\infty}(\mathbb{R}_{n}).$
(63)

Where
$f(x+2l)=f(x),\;l\in\mathbb{R}^{n},\;[-l,l]=\\{x\in\mathbb{R}^{n}|-l\leq{x}\leq{l}\\}$.

If $g(\partial_{x})=\partial_{x}^{\alpha},\;\forall\alpha\in\mathbb{N}^{n}$,
then (63) gives a definition of generalized derivative on $L^{2}([-l,l])$.

###  2.5 Integral representation of abstract operators

Clearly, such algorithms as (2.2)-(2.2) do not exist for most abstract
operators. Therefore, for more complex abstract operators, establishing the
integral expression is the fourth important way to extend the domain of
abstract operators. In reference [2] Guangqing Bi has obtained the following
results:

Theorem BI1. Let $P(\partial_{x})$ be an $m$-order partial differential
operator of any kind and there exist $a_{1},\,a_{2},\,\cdots,\,a_{k}$ of real
and partial differential operators $A_{1},\,A_{2},\,\cdots,\,A_{k}$ of the
order less than $[(m+1)/2]$ such that
$P(\partial_{x})\equiv{a}_{1}A_{1}^{2}+a_{2}A_{2}^{2}+\cdots+a_{k}A_{k}^{2}$.
If $k=2\nu+3,\;\nu=0,\,1,\,2,\cdots$, then
$\forall{f(x)}\in{C^{m\nu}}(\mathbb{R}^{n})$ we have

$\displaystyle\frac{\sinh\left(\alpha{t}P(\partial_{x})^{1/2}\right)}{\alpha
P(\partial_{x})^{1/2}}f(x)$ $\displaystyle=$ $\displaystyle
t\underbrace{\int^{t}_{0}tdt\cdots}_{\nu}\int^{t}_{0}\,tdt\frac{(\alpha^{2}P(\partial_{x}))^{\nu}}{2^{\nu+2}\pi^{\nu+1}}$
(64)
$\displaystyle\times\,\int^{\pi}_{-\pi}\underbrace{\int^{\pi}_{0}\cdots}_{k-2}\int^{\pi}_{0}e^{\eta_{1}a_{1}^{1/2}A_{1}+\cdots+\eta_{k}a_{k}^{1/2}A_{k}}f(x)\,d\sigma_{k}$
$\displaystyle+\,\sum^{\nu-1}_{i=0}\frac{t^{2i+1}}{(2i+1)!}(\alpha^{2}P(\partial_{x}))^{i}f(x).$

Where $\alpha$ is a real numbers,
$x\in\mathbb{R}^{n},\;t\in\mathbb{R}^{1}.\;\;\eta\in\mathbb{R}_{k}$ is the
integral variable and

$\displaystyle\eta_{1}$ $\displaystyle=$
$\displaystyle\alpha{t}\cos\theta_{1},$ $\displaystyle\eta_{2}$
$\displaystyle=$ $\displaystyle\alpha{t}\sin\theta_{1}\cos\theta_{2},$
$\displaystyle\cdots$ $\displaystyle\eta_{p}$ $\displaystyle=$
$\displaystyle\alpha{t}\sin\theta_{1}\sin\theta_{2}\cdots\sin\theta_{p-1}\cos\theta_{p},$
$\displaystyle\eta_{p+1}$ $\displaystyle=$
$\displaystyle\alpha{t}\sin\theta_{1}\sin\theta_{2}\cdots\sin\theta_{p}\cos\phi,$
$\displaystyle\eta_{p+2}$ $\displaystyle=$
$\displaystyle\eta_{k}\;=\;\alpha{t}\sin\theta_{1}\sin\theta_{2}\cdots\sin\theta_{p}\sin\phi;$
$d\sigma_{k}=\sin^{k-2}\theta_{1}\sin^{k-3}\theta_{2}\cdots\sin\theta_{k-2}d\theta_{1}d\theta_{2}\cdots{d}\theta_{k-2}d\phi.$

When $A_{1},\,\cdots,\,A_{k}$ in the right-hand of (64) are one order partial
differential operators, then the abstract operator
$e^{\eta_{1}a_{1}^{1/2}A_{1}+\cdots+\eta_{k}a_{k}^{1/2}A_{k}}$ is one of the
following five simplest operators:

$\exp(h\partial_{x}),\;\sin(h\partial_{x}),\;\cos(h\partial_{x}),\;\sinh(h\partial_{x})\;\mbox{and}\;\cosh(h\partial_{x}).$

Proof. In (64), let
$f(x)=e^{\xi{x}},\,x\in\mathbb{R}^{n},\xi\in\mathbb{R}_{n}$, and the symbol of
the partial differential operator $A_{j},j=1,2,\cdots,k$ is denoted by
$\chi_{j}(\xi),\;\beta_{j}=a_{j}^{1/2}\chi_{j}(\xi)$. By Definition 4, the
formula degenerates to its characteristic equation:

$\displaystyle\frac{\sinh\left(\alpha{t}P(\xi)^{1/2}\right)}{\alpha
P(\xi)^{1/2}}$ $\displaystyle=$ $\displaystyle
t\underbrace{\int^{t}_{0}tdt\cdots}_{\nu}\int^{t}_{0}\,tdt\frac{(\alpha^{2}P(\xi))^{\nu}}{2^{\nu+2}\pi^{\nu+1}}\int^{\pi}_{-\pi}\underbrace{\int^{\pi}_{0}\cdots}_{k-2}\int^{\pi}_{0}e^{\eta_{1}\beta_{1}+\cdots+\eta_{k}\beta_{k}}d\sigma_{k}$
(65)
$\displaystyle+\,\sum^{\nu-1}_{i=0}\frac{t^{2i+1}}{(2i+1)!}(\alpha^{2}P(\xi))^{i},\quad\nu=\frac{k-3}{2}.$

According to the analytic continuous fundamental theorem, we only need to
prove (65). Solving the integral on a hypersphere on the right side of (65),
we have

$\displaystyle\frac{\sinh\left(\alpha{t}P(\xi)^{1/2}\right)}{\alpha
P(\xi)^{1/2}}=t\underbrace{\int^{t}_{0}tdt\cdots}_{\nu}\int^{t}_{0}tdt\sum^{\infty}_{j=0}\frac{(\alpha^{2}P(\xi))^{\nu+j}t^{2j}}{(2j)!!(2j+2\nu+1)!!}+\sum^{\nu-1}_{i=0}\frac{t^{2i+1}(\alpha^{2}P(\xi))^{i}}{(2i+1)!}.$

Then it is proved by the termwise integration of the infinite series on the
right side of the equation.

Similarly, we have (See [1])

Theorem BI1’. Let $P(\partial_{x})$ be an $m$-order partial differential
operator of any kind and there exist $a_{1},\,a_{2},\,\cdots,\,a_{k}$ of real
and partial differential operators $A_{1},\,A_{2},\,\cdots,\,A_{k}$ of the
order less than $[(m+1)/2]$ such that
$P(\partial_{x})\equiv{a}_{1}A_{1}^{2}+a_{2}A_{2}^{2}+\cdots+a_{k}A_{k}^{2}$.
If $k=2\nu+3,\;\nu=0,\,1,\,2,\cdots$, then
$\forall{f(z)}\in{C^{\infty}}(\mathbb{C}^{n})$

$\displaystyle\frac{\sin\left({t}P(\partial_{x})^{1/2}\right)}{P(\partial_{x})^{1/2}}f(x)$
$\displaystyle=$ $\displaystyle
t\underbrace{\int^{t}_{0}tdt\cdots}_{\nu}\int^{t}_{0}tdt[-P(\partial_{x})]^{\nu}$
(66)
$\displaystyle\times\,\left(\frac{2}{\pi}\right)^{\nu+1}\underbrace{\int^{\pi/2}_{0}\cdots}_{k-1}\int^{\pi/2}_{0}\cos(\eta_{1}a_{1}^{1/2}A_{1})\cdots\cos(\eta_{k}a_{k}^{1/2}A_{k})f(x)\,d\sigma_{k}$
$\displaystyle+\,\sum^{\nu-1}_{i=0}\frac{t^{2i+1}}{(2i+1)!}[-P(\partial_{x})]^{i}f(x).$

Where $x\in\mathbb{R}^{n},\;t\in\mathbb{R}^{1}.\;\;\eta\in\mathbb{R}_{k}$ is
the integral variable and

$\displaystyle\eta_{1}$ $\displaystyle=$ $\displaystyle t\cos\theta_{1},$
$\displaystyle\eta_{2}$ $\displaystyle=$ $\displaystyle
t\sin\theta_{1}\cos\theta_{2},$ $\displaystyle\cdots$ $\displaystyle\eta_{p}$
$\displaystyle=$ $\displaystyle
t\sin\theta_{1}\sin\theta_{2}\cdots\sin\theta_{p-1}\cos\theta_{p},$
$\displaystyle\eta_{p+1}$ $\displaystyle=$ $\displaystyle
t\sin\theta_{1}\sin\theta_{2}\cdots\sin\theta_{p}\cos\phi,$
$\displaystyle\eta_{p+2}$ $\displaystyle=$
$\displaystyle\eta_{k}\;=\;t\sin\theta_{1}\sin\theta_{2}\cdots\sin\theta_{p}\sin\phi;$
$d\sigma_{k}=\sin^{k-2}\theta_{1}\sin^{k-3}\theta_{2}\cdots\sin\theta_{k-2}d\theta_{1}d\theta_{2}\cdots{d}\theta_{k-2}d\phi.$

If $A_{1},\,\cdots,\,A_{k}$ in (64) and (66) are partial differential
operators of the order great than 1, then the order can be lowered by taking
the following Theorem:

Theorem BI2. Let $P(\partial_{x}),\;x\in\mathbb{R}^{n}$ be a partial
differential operator of any order and $f(x)\in{C}(\mathbb{R}^{n})$ which make
the integral in the follow formula meaningful, then

$e^{\lambda
P(\partial_{x})}f(x)=\frac{1}{2\sqrt{\pi}}\int^{\infty}_{-\infty}e^{-\zeta^{2}/4}e^{\lambda^{1/2}\zeta
P(\partial_{x})^{1/2}}f(x)d\zeta,\quad\forall\lambda\in\mathbb{C}^{1}.$ (67)

Continuing the processing one can arrive at order one.

Example 6. By using the Theorem BI2 and Corollary 2 we have

$\exp\left(-a^{2}t\frac{\partial^{2}}{\partial{x}_{j}^{2}}\right)g(x)=\frac{1}{2\sqrt{\pi}}\int^{\infty}_{-\infty}e^{-\zeta^{2}/4}\cos\left(a\sqrt{t}\,\zeta\frac{\partial}{\partial{x_{j}}}\right)g(x)d\zeta,\quad{t}\in\overline{\mathbb{R}_{+}^{1}}.$
$\cos\left(\lambda\frac{\partial^{2}}{\partial{x}_{j}^{2}}\right)g(x)=\frac{1}{2\sqrt{\pi}}\int^{\infty}_{-\infty}e^{-\zeta^{2}/4}e^{\sqrt{\lambda/2}\,\zeta\frac{\partial}{\partial{x}_{j}}}\cos\left(\sqrt{\frac{\lambda}{2}}\,\zeta\frac{\partial}{\partial{x}_{j}}\right)g(x)d\zeta,$
$\sin\left(a\lambda\frac{\partial}{\partial{x}_{j}}+\lambda\frac{\partial^{2}}{\partial{x}_{j}^{2}}\right)g(x)=\frac{1}{2\sqrt{\pi}}\int^{\infty}_{-\infty}e^{-\zeta^{2}/4}e^{\sqrt{\lambda/2}\,\zeta\frac{\partial}{\partial{x}_{j}}}\sin\left(h_{\lambda,a}(\zeta)\frac{\partial}{\partial{x}_{j}}\right)g(x)d\zeta.$

Where
$h_{\lambda,a}(\zeta)=a\lambda+\sqrt{\lambda/2}\zeta,\;\forall\lambda,a\in\mathbb{C}^{1}$.

With the expansion of definitions of abstract operators, it is necessary to
add new algorithms continuously.

##  3 Abstract operators and linear higher-order partial differential
equations

###  3.1 Applications of Laplace transform for n-dimensional space

Solving the ordinary or partial differential equations, is constructing the
algorithms of the inverse operators of ordinary or partial differential
operators. In terms of abstract operators, solving the initial value problem
of partial differential equations for n-dimensional space is similar to
solving the ordinary differential equations with respect to the variable $t$,
thus we can introduce the Laplace transform to further simplify the solving
process.

Corollary 7. Let $g(\int^{t}_{0}\\!\cdot\,{t}dt)$ be the abstract operators
taking $\int^{t}_{0}\\!\cdot\,{t}dt$ as the operator element, and
$g(-\frac{1}{s}\frac{\partial}{\partial{s}})$ be the abstract operator taking
$-\frac{1}{s}\frac{\partial}{\partial{s}}$ as the operator element, and the
symbols of these two abstract operators are the same, then for any Laplace
transformable function $f(t)$ we have

$g\left(\int^{t}_{0}\\!\cdot\,{t}dt\right)f(t)=\mathscr{L}^{-1}\,g\left(-\frac{1}{s}\frac{\partial}{\partial{s}}\right)\mathscr{L}f(t),\quad{t}\in\overline{\mathbb{R}_{+}^{1}}.$
(68)

Where $\mathscr{L}$ and $\mathscr{L}^{-1}$ are the Laplace transform operator
and its inverse operator respectively, and $\mathscr{L}f(t)=F(s)$.

Proof. According to the properties of Laplace transform, we generally have

$\left(\int^{t}_{0}\\!\cdot\,{t}dt\right)^{m}f(t)=\mathscr{L}^{-1}\,\left(-\frac{1}{s}\frac{\partial}{\partial{s}}\right)^{m}F(s),\quad\forall{m}\in\mathbb{N}.$

According to the abstract operator fundamental theorem, we get the Corollary
7.

Similarly we have

$g\left(\int_{0}^{t}\\!\cdot\,dt\right)f(t)=\mathscr{L}^{-1}g\left(\frac{1}{s}\right)\mathscr{L}f(t),\quad{t}\in\overline{\mathbb{R}_{+}^{1}}.$
(69)

Therefore, using properties of the Laplace transform and the abstract operator
fundamental theorem, we respectively obtain the abstract operators taking
$\int^{t}_{0}\\!\cdot\,{t}dt$ and $\int^{t}_{0}\\!\cdot\,dt$ as the operator
elements. This type of formulas can also be applied in solving certain
integral equations.

Theorem 7. Let $m\in\mathbb{N}_{+}$, $P(\partial_{x})$ be arbitrary order
partial differential equations for n-dimensional space. Then we have

$\left\\{\begin{array}[]{l@{\qquad}l}\displaystyle\left(\frac{\partial^{2}}{\partial{t^{2}}}-P(\partial_{x})\right)^{m}u=f(x,t),&x\in\mathbb{R}^{n},\;t\in\overline{\mathbb{R}_{+}^{1}},\;\forall{m}\in\mathbb{N}_{+},\\\
\displaystyle\left.\frac{\partial^{j}u}{\partial{t^{j}}}\right|_{t=0}=\varphi_{j}(x),&j=0,1,2,\ldots,2m-1.\end{array}\right.$
(70) $\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\int^{t}_{0}\int^{t-\tau}_{0}\frac{\left[(t-\tau)^{2}-\tau^{\prime
2}\right]^{m-2}}{(2m-2)!!\,(2m-4)!!}\,\frac{\sinh\left(\tau^{\prime}P(\partial_{x})^{1/2}\right)}{P(\partial_{x})^{1/2}}\,f(x,\tau)\,\tau^{\prime}d\tau^{\prime}\,d\tau$
(71)
$\displaystyle+\,\sum^{m-1}_{k=0}(-1)^{k}{m\choose{k}}P(\partial_{x})^{k}\sum^{2m-1-2k}_{j=0}\frac{\partial^{2m-1-2k-j}}{\partial
t^{2m-1-2k-j}}\int^{t}_{0}\frac{(t^{2}-\tau^{2})^{m-2}\tau}{(2m-2)!!\,(2m-4)!!}$
$\displaystyle\times\,\frac{\sinh\left(\tau
P(\partial_{x})^{1/2}\right)}{P(\partial_{x})^{1/2}}\,\varphi_{j}(x)\,d\tau.$

Proof. The Laplace transform of the partial differential equation, with
respect to $t$ and considering the initial condition, is

$\sum^{m}_{k=0}(-1)^{k}{m\choose{k}}P(\partial_{x})^{k}\left(s^{2m-2k}U(x,s)-\sum^{2m-1-2k}_{j=0}s^{2m-1-2k-j}\varphi_{j}(x)\right)=F(x,s).$

Where $U(x,s)=\mathscr{L}u(x,t),\;F(x,s)=\mathscr{L}f(x,t)$. Let
$G_{m}(\partial_{x},t)=\mathscr{L}^{-1}[1/(s^{2}-P(\partial_{x}))^{m}]$,
solving $U(x,s)$ and its inverse Laplace transform is

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\mathscr{L}^{-1}U(x,s)=\mathscr{L}^{-1}\frac{1}{(s^{2}-P(\partial_{x}))^{m}}F(x,s)$
$\displaystyle+\,\mathscr{L}^{-1}\sum^{m-1}_{k=0}(-1)^{k}{m\choose{k}}P(\partial_{x})^{k}\sum^{2m-1-2k}_{j=0}\frac{s^{2m-1-2k-j}}{(s^{2}-P(\partial_{x}))^{m}}\varphi_{j}(x)$
$\displaystyle=$ $\displaystyle G_{m}(\partial_{x},t)\ast f(x,t)$
$\displaystyle+\,\sum^{m-1}_{k=0}(-1)^{k}{m\choose{k}}P(\partial_{x})^{k}\sum^{2m-1-2k}_{j=0}\frac{\partial^{2m-1-2k-j}}{\partial
t^{2m-1-2k-j}}G_{m}(\partial_{x},t)\varphi_{j}(x).$

Now let us solve $G_{m}(\partial_{x},t)$. In Corollary 7, taking
$g(\xi)=\xi^{m-1},\;\xi\in\mathbb{R}_{1}$ as the symbol of the abstract
operator, and let $f(t)=\sin{bt},\;t\in\mathbb{R}^{1}$, we have

$\left(\int^{t}_{0}\cdot\,tdt\right)^{m-1}\\!\\!\sin{bt}=\mathscr{L}^{-1}\left(-\frac{1}{s}\frac{\partial}{\partial{s}}\right)^{m-1}\\!\\!\frac{b}{s^{2}+b^{2}}=\mathscr{L}^{-1}\frac{2^{m-1}(m-1)!}{(s^{2}+b^{2})^{m}}b.$

Let $b=iP(\xi)^{1/2},\;\xi\in\mathbb{R}_{n}$, we have

$\mathscr{L}^{-1}\frac{1}{(s^{2}-P(\xi))^{m}}=\frac{1}{(2m-2)!!}\left(\int^{t}_{0}\cdot\,tdt\right)^{m-1}\frac{\sinh\left(tP(\xi)^{1/2}\right)}{P(\xi)^{1/2}}.$

Taking this one as the characteristic equation, according to the analytic
continuous fundamental theorem, we have

$G_{m}(\partial_{x},t)=\mathscr{L}^{-1}\frac{1}{(s^{2}-P(\partial_{x}))^{m}}=\frac{1}{(2m-2)!!}\left(\int^{t}_{0}\cdot\,tdt\right)^{m-1}\frac{\sinh\left(tP(\partial_{x})^{1/2}\right)}{P(\partial_{x})^{1/2}}.$
(72)

By using (8) in reference [3], we can easily derive the following integral
formula

$\left(\int^{x}_{a}\cdot\,xdx\right)^{m}f(x)=\underbrace{\int^{x}_{a}xdx\cdots}_{m}\int^{x}_{a}f(x)\,xdx=\int^{x}_{a}\frac{(x^{2}-\xi^{2})^{m-1}}{(2m-2)!!}f(\xi)\,\xi
d\xi.$ (73)

Applying (73) to (72), we have the expression of abstract operator
$G_{m}(\partial_{x},t)$:

$G_{m}(\partial_{x},t)g(x)=\int^{t}_{0}\frac{(t^{2}-\tau^{\prime
2})^{m-2}}{(2m-2)!!\,(2m-4)!!}\,\frac{\sinh\left(\tau^{\prime}P(\partial_{x})^{1/2}\right)}{P(\partial_{x})^{1/2}}g(x)\,\tau^{\prime}d\tau^{\prime}.$
(74)

Thus Theorem 7 is proved.

In reference [2] the Guangqing Bi has obtained the following results:

Theorem BI3. Let $a_{1},a_{2},\ldots,a_{m}$ be arbitrary real or complex
numbers different from each other, $P(\partial_{x})$ be a partial differential
operator of any order, then we have

$\left\\{\begin{array}[]{l@{\qquad}l}\displaystyle\prod^{m}_{i=1}(\frac{\partial}{\partial{t}}-a_{i}P(\partial_{x}))u=f(x,t),&x\in\mathbb{R}^{n},\;t\in\overline{\mathbb{R}_{+}^{1}},\;\forall{m}\in\mathbb{N}_{+},\\\
\displaystyle\left.\frac{\partial^{j}u}{\partial{t^{j}}}\right|_{t=0}=0,&j=0,1,2,\ldots,m-1.\end{array}\right.$
(75)
$u(x,t)=\int^{t}_{0}\int^{t-\tau}_{0}\frac{(t-\tau-\tau^{\prime})^{m-2}}{(m-2)!}\sum^{m}_{j=1}\frac{a_{j}^{m-1}}{\prod^{m}_{i=1\atop
i\neq{j}}(a_{j}-a_{i})}e^{\tau^{\prime}a_{j}P(\partial_{x})}f(x,\tau)\,d\tau^{\prime}d\tau.$
(76)

Theorem BI4. Let $a_{1},a_{2},\ldots,a_{m}$ be arbitrary real or complex
numbers different from each other, $P(\partial_{x})$ be a partial differential
operator of any order, then we have

$\left\\{\begin{array}[]{l@{\qquad}l}\displaystyle\prod^{m}_{i=1}(\frac{\partial^{2}}{\partial{t^{2}}}-a_{i}^{2}P(\partial_{x}))u=f(x,t),&x\in\mathbb{R}^{n},\;t\in\overline{\mathbb{R}_{+}^{1}},\;\forall{m}\in\mathbb{N}_{+},\\\
\displaystyle\left.\frac{\partial^{j}u}{\partial{t^{j}}}\right|_{t=0}=0,&j=0,1,2,\ldots,2m-1.\end{array}\right.$
(77)
$u(x,t)=\int^{t}_{0}\int^{t-\tau}_{0}\frac{(t-\tau-\tau^{\prime})^{2m-3}}{(2m-3)!}\sum^{m}_{j=1}\frac{a_{j}^{2m-2}}{\prod^{m}_{i=1\atop
i\neq{j}}(a_{j}^{2}-a_{i}^{2})}\frac{\sinh(\tau^{\prime}a_{j}P(\partial_{x})^{1/2})}{a_{j}P(\partial_{x})^{1/2}}f(x,\tau)\,d\tau^{\prime}d\tau.$
(78)

On this basis, by using the abstract operators and Laplace transform we have
obtained the following results:

Theorem 8. Let $a_{1},a_{2},\ldots,a_{m}$ be arbitrary real or complex roots
different from each other for
$b_{0}+b_{1}\chi+b_{2}\chi^{2}+\cdots+b_{m}\chi^{m}=0$, and
$P(\partial_{x},\partial_{t})$ be a partial differential operator defined by

$P(\partial_{x},\partial_{t})=\sum^{m}_{k=0}b_{k}P(\partial_{x})^{m-k}\frac{\partial^{k}}{\partial{t^{k}}},\quad{x}\in\mathbb{R}^{n},\;t\in\overline{\mathbb{R}_{+}^{1}},\;\forall{m}\in\mathbb{N}_{+}.$

Where $P(\partial_{x})$ is a partial differential operator of any order. Then
we have

$\left\\{\begin{array}[]{l@{\qquad}l}\displaystyle
P(\partial_{x},\partial_{t})u=f(x,t),&x\in\mathbb{R}^{n},\;t\in\overline{\mathbb{R}_{+}^{1}},\;\forall{m}\in\mathbb{N}_{+},\\\
\displaystyle\left.\frac{\partial^{r}u}{\partial{t^{r}}}\right|_{t=0}=\varphi_{r}(x),&r=0,1,2,\ldots,m-1.\end{array}\right.$
(79) $\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\int^{t}_{0}\int^{t-\tau}_{0}\frac{(t-\tau-\tau^{\prime})^{m-2}}{(m-2)!}\sum^{m}_{j=1}\frac{a_{j}^{m-1}}{\prod^{m}_{i=1\atop
i\neq{j}}(a_{j}-a_{i})}\,e^{\tau^{\prime}a_{j}P(\partial_{x})}f(x,\tau)\,d\tau^{\prime}d\tau$
(80)
$\displaystyle+\,\sum^{m}_{k=1}b_{k}P(\partial_{x})^{m-k}\sum^{k-1}_{r=0}\frac{\partial^{k-1-r}}{\partial
t^{k-1-r}}\int^{t}_{0}\frac{(t-\tau)^{m-2}}{(m-2)!}$
$\displaystyle\times\,\sum^{m}_{j=1}\frac{a_{j}^{m-1}}{\prod^{m}_{i=1\atop
i\neq{j}}(a_{j}-a_{i})}\,e^{\tau{a_{j}}P(\partial_{x})}\,\varphi_{r}(x)\,d\tau.$

Theorem 9. Let $a_{1},a_{2},\ldots,a_{m}$ be arbitrary real or complex roots
different from each other, satisfy
$\sum^{m}_{k=0}b_{2k}\chi^{2k}=\prod^{m}_{i=1}(\chi^{2}-a_{i}^{2})$, and
$P(\partial_{x},\partial_{t})$ be a partial differential operators defined by

$P(\partial_{x},\partial_{t})=\sum^{m}_{k=0}b_{2k}P(\partial_{x})^{m-k}\frac{\partial^{2k}}{\partial{t^{2k}}},\quad{x}\in\mathbb{R}^{n},\;t\in\overline{\mathbb{R}_{+}^{1}},\;\forall{m}\in\mathbb{N}_{+}.$

Where $P(\partial_{x})$ be a partial differential operator of any order, then
we have

$\left\\{\begin{array}[]{l@{\qquad}l}\displaystyle
P(\partial_{x},\partial_{t})u=f(x,t),&x\in\mathbb{R}^{n},\;t\in\overline{\mathbb{R}_{+}^{1}},\;\forall{m}\in\mathbb{N}_{+},\\\
\displaystyle\left.\frac{\partial^{r}u}{\partial{t^{r}}}\right|_{t=0}=\varphi_{r}(x),&r=0,1,2,\ldots,2m-1.\end{array}\right.$
(81) $\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\int^{t}_{0}\int^{t-\tau}_{0}\frac{(t-\tau-\tau^{\prime})^{2m-3}}{(2m-3)!}\sum^{m}_{j=1}\frac{a_{j}^{2m-2}}{\prod^{m}_{i=1\atop
i\neq{j}}(a_{j}^{2}-a_{i}^{2})}\frac{\sinh(\tau^{\prime}a_{j}P(\partial_{x})^{1/2})}{a_{j}P(\partial_{x})^{1/2}}f(x,\tau)\,d\tau^{\prime}d\tau$
(82)
$\displaystyle+\,\sum^{m}_{k=1}b_{2k}P(\partial_{x})^{m-k}\sum^{2k-1}_{r=0}\frac{\partial^{2k-1-r}}{\partial
t^{2k-1-r}}\int^{t}_{0}\frac{(t-\tau)^{2m-3}}{(2m-3)!}$
$\displaystyle\times\,\sum^{m}_{j=1}\frac{a_{j}^{2m-2}}{\prod^{m}_{i=1\atop
i\neq{j}}(a_{j}^{2}-a_{i}^{2})}\frac{\sinh(\tau{a_{j}}P(\partial_{x})^{1/2})}{a_{j}P(\partial_{x})^{1/2}}\,\varphi_{r}(x)\,d\tau.$

Let us prove these two theorems. According to the Theorem BI3 and Theorem BI4,
we just need to prove the following Corollary of Theorem 8 and Theorem 9:

Corollary 8. Let $a_{1},a_{2},\ldots,a_{m}$ be arbitrary real or complex roots
different from each other for
$b_{0}+b_{1}\chi+b_{2}\chi^{2}+\cdots+b_{m}\chi^{m}=0$, and
$P(\partial_{x},\partial_{t})$ be a partial differential operators defined by

$P(\partial_{x},\partial_{t})=\sum^{m}_{k=0}b_{k}P(\partial_{x})^{m-k}\frac{\partial^{k}}{\partial{t^{k}}},\quad{x}\in\mathbb{R}^{n},\;t\in\overline{\mathbb{R}_{+}^{1}},\;\forall{m}\in\mathbb{N}_{+}.$

Where $P(\partial_{x})$ be a partial differential operator of any order, then
we have

$\left\\{\begin{array}[]{l@{\qquad}l}\displaystyle
P(\partial_{x},\partial_{t})u=0,&x\in\mathbb{R}^{n},\;t\in\overline{\mathbb{R}_{+}^{1}},\;\forall{m}\in\mathbb{N}_{+},\\\
\displaystyle\left.\frac{\partial^{r}u}{\partial{t^{r}}}\right|_{t=0}=\varphi_{r}(x),&r=0,1,2,\ldots,m-1.\end{array}\right.$
(83) $\displaystyle
u(x,t)=\sum^{m}_{k=1}b_{k}P(\partial_{x})^{m-k}\sum^{k-1}_{r=0}\frac{\partial^{k-1-r}}{\partial
t^{k-1-r}}\int^{t}_{0}\frac{(t-\tau)^{m-2}}{(m-2)!}\sum^{m}_{j=1}\frac{a_{j}^{m-1}e^{\tau{a_{j}}P(\partial_{x})}}{\prod^{m}_{i=1\atop
i\neq{j}}(a_{j}-a_{i})}\,\varphi_{r}(x)\,d\tau.$ (84)

Proof. Considering initial conditions, the Laplace transform of the Eq (83)
with respect to $t$ is

$\sum^{m}_{k=0}b_{k}P(\partial_{x})^{m-k}\left(s^{k}U(x,s)-\sum^{k-1}_{r=0}s^{k-1-r}\varphi_{r}(x)\right)=0.$

Where $U(x,s)=\mathscr{L}u(x,t)$, considering
$\prod^{m}_{i=1}(s-a_{i}P(\partial_{x}))=\sum^{m}_{k=0}b_{k}s^{k}P(\partial_{x})^{m-k}$
we have

$\prod^{m}_{i=1}(s-a_{i}P(\partial_{x}))U(x,s)-\sum^{m}_{k=1}b_{k}P(\partial_{x})^{m-k}\sum^{k-1}_{r=0}s^{k-1-r}\varphi_{r}(x)=0.$

We need to introduce an abstract operators $G_{m}(\partial_{x},t)$, defined as

$G_{m}(\partial_{x},t)=\mathscr{L}^{-1}\frac{1}{\prod^{m}_{i=1}(s-a_{i}P(\partial_{x}))}.$

By solving $U(x,s)$, we have its inverse Laplace transform:

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\mathscr{L}^{-1}U(x,s)=\sum^{m}_{k=1}b_{k}P(\partial_{x})^{m-k}\sum^{k-1}_{r=0}\mathscr{L}^{-1}\frac{s^{k-1-r}}{\prod^{m}_{i=1}(s-a_{i}P(\partial_{x}))}\,\varphi_{r}(x)$
(85) $\displaystyle=$
$\displaystyle\sum^{m}_{k=1}b_{k}P(\partial_{x})^{m-k}\sum^{k-1}_{r=0}\frac{\partial^{k-1-r}}{\partial{t^{k-1-r}}}\,G_{m}(\partial_{x},t)\varphi_{r}(x).$

Now let us solve $G_{m}(\partial_{x},t)$. Considering initial conditions, the
Laplace transform of the Eq (75) with respect to $t$ is

$\prod^{m}_{i=1}(s-a_{i}P(\partial_{x}))U(x,s)=F(x,s),$

where $F(x,s)=\mathscr{L}f(x,t)$. By solving $U(x,s)$ and using the
convolution theorem, we have its inverse Laplace transform:

$u(x,t)=\mathscr{L}^{-1}U(x,s)=\mathscr{L}^{-1}\frac{1}{\prod^{m}_{i=1}(s-a_{i}P(\partial_{x}))}F(x,s)=G_{m}(\partial_{x},t)*f(x,t).$

By comparing (76) with $u(x,t)=G_{m}(\partial_{x},t)*f(x,t)$, we have the
expression of the abstract operators $G_{m}(\partial_{x},t)$:

$G_{m}(\partial_{x},t)=\int^{t}_{0}\frac{(t-\tau)^{m-2}}{(m-2)!}\sum^{m}_{j=1}\frac{a_{j}^{m-1}}{\prod^{m}_{i=1\atop
i\neq{j}}(a_{j}-a_{i})}e^{\tau{a_{j}}P(\partial_{x})}d\tau.$ (86)

Applying (86) to (85), thus the Corollary 8 is proved.

Corollary 9. Let $a_{1},a_{2},\ldots,a_{m}$ be arbitrary real or complex roots
different from each other, which satisfy
$\sum^{m}_{k=0}b_{2k}\chi^{2k}=\prod^{m}_{i=1}(\chi^{2}-a_{i}^{2})$, and
$P(\partial_{x},\partial_{t})$ be a partial differential operator defined by

$P(\partial_{x},\partial_{t})=\sum^{m}_{k=0}b_{2k}P(\partial_{x})^{m-k}\frac{\partial^{2k}}{\partial{t^{2k}}},\quad{x}\in\mathbb{R}^{n},\;t\in\overline{\mathbb{R}_{+}^{1}},\;\forall{m}\in\mathbb{N}_{+}.$

Where $P(\partial_{x})$ is a partial differential operator of any order, then
we have

$\left\\{\begin{array}[]{l@{\qquad}l}\displaystyle
P(\partial_{x},\partial_{t})u=0,&x\in\mathbb{R}^{n},\;t\in\overline{\mathbb{R}_{+}^{1}},\;\forall{m}\in\mathbb{N}_{+},\\\
\displaystyle\left.\frac{\partial^{r}u}{\partial{t^{r}}}\right|_{t=0}=\varphi_{r}(x),&r=0,1,2,\ldots,2m-1.\end{array}\right.$
(87) $\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\sum^{m}_{k=1}b_{2k}P(\partial_{x})^{m-k}\sum^{2k-1}_{r=0}\frac{\partial^{2k-1-r}}{\partial
t^{2k-1-r}}\int^{t}_{0}\frac{(t-\tau)^{2m-3}}{(2m-3)!}$ (88)
$\displaystyle\times\,\sum^{m}_{j=1}\frac{a_{j}^{2m-2}}{\prod^{m}_{i=1\atop
i\neq{j}}(a_{j}^{2}-a_{i}^{2})}\frac{\sinh(\tau{a_{j}}P(\partial_{x})^{1/2})}{a_{j}P(\partial_{x})^{1/2}}\,\varphi_{r}(x)\,d\tau.$

Proof. Considering initial conditions, the Laplace transform of the Eq (87)
with respect to $t$ is

$\sum^{m}_{k=0}b_{2k}P(\partial_{x})^{m-k}\left(s^{2k}U(x,s)-\sum^{2k-1}_{r=0}s^{2k-1-r}\varphi_{r}(x)\right)=0.$

Where $U(x,s)=\mathscr{L}u(x,t)$, considering
$\prod^{m}_{i=1}(s^{2}-a_{i}^{2}P(\partial_{x}))=\sum^{m}_{k=0}b_{2k}s^{2k}P(\partial_{x})^{m-k}$
we have

$\prod^{m}_{i=1}(s^{2}-a_{i}^{2}P(\partial_{x}))U(x,s)-\sum^{m}_{k=1}b_{2k}P(\partial_{x})^{m-k}\sum^{2k-1}_{r=0}s^{2k-1-r}\varphi_{r}(x)=0.$

We need to introduce an abstract operators $G_{m}(\partial_{x},t)$, defined as

$G_{m}(\partial_{x},t)=\mathscr{L}^{-1}\frac{1}{\prod^{m}_{i=1}(s^{2}-a_{i}^{2}P(\partial_{x}))}.$

By solving $U(x,s)$, we have its inverse Laplace transform:

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\mathscr{L}^{-1}U(x,s)=\sum^{m}_{k=1}b_{2k}P(\partial_{x})^{m-k}\sum^{2k-1}_{r=0}\mathscr{L}^{-1}\frac{s^{2k-1-r}}{\prod^{m}_{i=1}(s^{2}-a_{i}^{2}P(\partial_{x}))}\,\varphi_{r}(x)$
(89) $\displaystyle=$
$\displaystyle\sum^{m}_{k=1}b_{2k}P(\partial_{x})^{m-k}\sum^{2k-1}_{r=0}\frac{\partial^{2k-1-r}}{\partial{t^{2k-1-r}}}\,G_{m}(\partial_{x},t)\varphi_{r}(x).$

Now let us solve the $G_{m}(\partial_{x},t)$. Considering initial conditions,
the Laplace transform of the Eq (77) with respect to $t$ is

$\prod^{m}_{i=1}(s^{2}-a_{i}^{2}P(\partial_{x}))U(x,s)=F(x,s),$

where $F(x,s)=\mathscr{L}f(x,t)$. By solving $U(x,s)$ and using the
convolution theorem, we have its inverse Laplace transform:

$u(x,t)=\mathscr{L}^{-1}U(x,s)=\mathscr{L}^{-1}\frac{1}{\prod^{m}_{i=1}(s^{2}-a_{i}^{2}P(\partial_{x}))}F(x,s)=G_{m}(\partial_{x},t)*f(x,t).$

By comparing (78) with $u(x,t)=G_{m}(\partial_{x},t)*f(x,t)$, we have the
expression of the abstract operators $G_{m}(\partial_{x},t)$:

$G_{m}(\partial_{x},t)=\int^{t}_{0}\frac{(t-\tau)^{2m-3}}{(2m-3)!}\sum^{m}_{j=1}\frac{a_{j}^{2m-2}}{\prod^{m}_{i=1\atop
i\neq{j}}(a_{j}^{2}-a_{i}^{2})}\frac{\sinh(\tau{a_{j}}P(\partial_{x})^{1/2})}{a_{j}P(\partial_{x})^{1/2}}\,d\tau.$
(90)

Applying (90) to (89), thus the Corollary 9 is proved.

###  3.2 Analytic solution of Cauchy problem

Theorem 10. Let
$\Delta_{n}=\frac{\partial^{2}}{\partial{x_{1}}^{2}}+\frac{\partial^{2}}{\partial{x_{2}}^{2}}+\cdots+\frac{\partial^{2}}{\partial{x_{n}}^{2}}$
be an n-dimensional Laplacian. If $n-2=2\nu+1,\;\nu\in\mathbb{N}$, then
$\forall{f(x,\tau)}\in{C^{2\nu}}(\mathbb{R}^{n})$, and
$\forall\varphi_{j}(x)\in{C}^{2m+2\nu-2}(\mathbb{R}^{n})$, we have

$\left\\{\begin{array}[]{l@{\qquad}l}\displaystyle\left(\frac{\partial^{2}}{\partial{t^{2}}}-a^{2}\Delta_{n}\right)^{m}u=f(x,t),&x\in\mathbb{R}^{n},\;t\in\overline{\mathbb{R}_{+}^{1}},\;\forall{m}\in\mathbb{N}_{+},\\\
\displaystyle\left.\frac{\partial^{j}u}{\partial{t^{j}}}\right|_{t=0}=\varphi_{j}(x),&j=0,1,2,\ldots,2m-1.\end{array}\right.$
(91) $\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\int^{t}_{0}d\tau\int^{t-\tau}_{0}d\tau^{\prime}\frac{\left[(t-\tau)^{2}-\tau^{\prime
2}\right]^{m-2}\tau^{\prime 2}}{(2m-2)!!\,(2m-4)!!}$ (92)
$\displaystyle\times\underbrace{\int^{\tau^{\prime}}_{0}\tau^{\prime}d\tau^{\prime}\cdots}_{\nu}\int^{\tau^{\prime}}_{0}\frac{(a^{2}\Delta_{n})^{\nu}}{S^{\prime}_{n}}\int_{S^{\prime}_{n}}f(\xi^{\prime},\tau)\,dS^{\prime}_{n}\,\tau^{\prime}d\tau^{\prime}$
$\displaystyle+\,\frac{1}{(2m-2)!!}\sum^{\nu-1}_{r=0}\int^{t}_{0}\frac{(t-\tau)^{2m+2r-1}}{(2m+2r-1)!!\,(2r)!!}(a^{2}\Delta_{n})^{r}f(x,\tau)\,d\tau$
$\displaystyle+\,\sum^{m-1}_{k=0}(-1)^{k}{m\choose{k}}(a^{2}\Delta_{n})^{k+\nu}\sum^{2m-1-2k}_{j=0}\frac{\partial^{2m-1-2k-j}}{\partial{t}^{2m-1-2k-j}}\int^{t}_{0}d\tau\frac{(t^{2}-\tau^{2})^{m-2}\tau^{2}}{(2m-2)!!\,(2m-4)!!}$
$\displaystyle\times\underbrace{\int^{\tau}_{0}\tau{d}\tau\cdots}_{\nu}\int^{\tau}_{0}\frac{1}{S_{n}}\int_{S_{n}}\varphi_{j}(\xi)\,dS_{n}\,\tau{d}\tau+\sum^{m-1}_{k=0}(-1)^{k}{m\choose{k}}\sum^{2m-1-2k}_{j=0}$
$\displaystyle\times\sum^{\nu-1}_{i=0}{m-1+i\choose{i}}\frac{t^{2k+2i+j}}{(2k+2i+j)!}(a^{2}\Delta_{n})^{k+i}\varphi_{j}(x).$

Where $n-2=2\nu+1$, $S^{\prime}_{n}=2(2\pi)^{\nu+1}(a\tau^{\prime})^{n-1}$,
$S_{n}=2(2\pi)^{\nu+1}(a\tau)^{n-1}$, and $\xi^{\prime}\in\mathbb{R}_{n}$ is
the integral variable. The integral is on the hypersphere
$(\xi^{\prime}_{1}-x_{1})^{2}+(\xi^{\prime}_{2}-x_{2})^{2}+\cdots+(\xi^{\prime}_{n}-x_{n})^{2}=(a\tau^{\prime})^{2}$,
and $dS^{\prime}_{n}$ is its surface element. $\xi\in\mathbb{R}_{n}$ is the
integral variable on the hypersphere
$(\xi_{1}-x_{1})^{2}+(\xi_{2}-x_{2})^{2}+\cdots+(\xi_{n}-x_{n})^{2}=(a\tau)^{2}$,
and $dS_{n}$ is its surface element.

Proof. According to (64), we can easily derive:

$\displaystyle\frac{\sinh\left(at\Delta_{n}{}^{1/2}\right)}{a\Delta_{n}{}^{1/2}}f(x)$
$\displaystyle=$ $\displaystyle
t\underbrace{\int^{t}_{0}tdt\cdots}_{\nu}\int^{t}_{0}\frac{(a^{2}\Delta_{n})^{\nu}}{S_{n}}\int_{S_{n}}f(\xi)\,dS_{n}\,tdt$
(93)
$\displaystyle+\,\sum^{\nu-1}_{i=0}\frac{t^{2i+1}}{(2i+1)!}(a^{2}\Delta_{n})^{i}f(x),\quad\forall{f(x)}\in{C}^{2\nu}(\mathbb{R}^{n}).$

Where $\Delta_{n}$ is an n-dimensional Laplacian, $n-2=2\nu+1$,
$S_{n}=2(2\pi)^{\nu+1}(at)^{n-1}$. $\xi\in\mathbb{R}_{n}$ is the integral
variable on the hypersphere
$(\xi_{1}-x_{1})^{2}+(\xi_{2}-x_{2})^{2}+\cdots+(\xi_{n}-x_{n})^{2}=(at)^{2}$,
and $dS_{n}$ is its surface element.

In Theorem 7, let $P(\partial_{x})=a^{2}\Delta_{n}$, then Theorem 10 is proved
by the substitution of (93).

Theorem 11. Let $a_{1},a_{2},\ldots,a_{m}$ be arbitrary real roots different
from each other, which satisfy
$\sum^{m}_{k=0}b_{2k}\chi^{2k}=\prod^{m}_{i=1}(\chi^{2}-a_{i}^{2})$, and
$P(\partial_{x},\partial_{t})$ be a partial differential operator defined by

$P(\partial_{x},\partial_{t})=\sum^{m}_{k=0}b_{2k}\Delta_{n}^{m-k}\frac{\partial^{2k}}{\partial{t^{2k}}},\quad{x}\in\mathbb{R}^{n},\;t\in\overline{\mathbb{R}_{+}^{1}},\;\forall{m}\in\mathbb{N}_{+}.$

Where $\Delta_{n}$ is an n-dimensional Laplacian. If
$n-2=2\nu+1,\;\nu=0,1,2,\cdots$, then
$\forall\varphi_{r}(x)\in{C^{2m+2\nu-2}}(\mathbb{R}^{n})$, and
$\forall{f(x,\tau)}\in{C^{2\nu}}(\mathbb{R}^{n})$, we have

$\left\\{\begin{array}[]{l@{\qquad}l}\displaystyle
P(\partial_{x},\partial_{t})u=f(x,t),&x\in\mathbb{R}^{n},\;t\in\overline{\mathbb{R}_{+}^{1}},\;\forall{m}\in\mathbb{N}_{+},\\\
\displaystyle\left.\frac{\partial^{r}u}{\partial{t^{r}}}\right|_{t=0}=\varphi_{r}(x),&r=0,1,2,\ldots,2m-1.\end{array}\right.$
(94) $\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\int^{t}_{0}d\tau\int^{t-\tau}_{0}d\tau^{\prime}\frac{(t-\tau-\tau^{\prime})^{2m-3}}{(2m-3)!}\sum^{m}_{j=1}\frac{a_{j}^{2m-2}}{\prod^{m}_{i=1\atop
i\neq{j}}(a_{j}^{2}-a_{i}^{2})}$ (95)
$\displaystyle\times\left(\tau^{\prime}\underbrace{\int^{\tau^{\prime}}_{0}\tau^{\prime}d\tau^{\prime}\cdots}_{\nu}\int^{\tau^{\prime}}_{0}\frac{(a_{j}^{2}\Delta_{n})^{\nu}}{S^{\prime}_{n,j}}\int_{S^{\prime}_{n,j}}f(\xi^{\prime},\tau)\,dS^{\prime}_{n,j}\,\tau^{\prime}d\tau^{\prime}+\sum^{\nu-1}_{l=0}\frac{a_{j}^{2l}\tau^{\prime
2l+1}}{(2l+1)!}\Delta_{n}^{l}f(x,\tau)\right)$
$\displaystyle+\,\sum^{m}_{k=1}b_{2k}\Delta_{n}^{m-k}\sum^{2k-1}_{r=0}\frac{\partial^{2k-1-r}}{\partial
t^{2k-1-r}}\int^{t}_{0}d\tau\frac{(t-\tau)^{2m-3}}{(2m-3)!}\sum^{m}_{j=1}\frac{a_{j}^{2m-2}}{\prod^{m}_{i=1\atop
i\neq{j}}(a_{j}^{2}-a_{i}^{2})}$
$\displaystyle\times\left(\tau\underbrace{\int^{\tau}_{0}\tau{d\tau}\cdots}_{\nu}\int^{\tau}_{0}\frac{(a_{j}^{2}\Delta_{n})^{\nu}}{S_{n,j}}\int_{S_{n,j}}\varphi_{r}(\xi)\,dS_{n,j}\,\tau{d\tau}+\sum^{\nu-1}_{l=0}\frac{a_{j}^{2l}\tau^{2l+1}}{(2l+1)!}\Delta_{n}^{l}\varphi_{r}(x)\right).$

Where $S^{\prime}_{n,j}=2(2\pi)^{\nu+1}(a_{j}\tau^{\prime})^{n-1}$,
$S_{n,j}=2(2\pi)^{\nu+1}(a_{j}\tau)^{n-1}$, and
$\xi^{\prime}\in\mathbb{R}_{n}$ is the integral variable. The integral is on
the hypersphere
$(\xi^{\prime}_{1}-x_{1})^{2}+(\xi^{\prime}_{2}-x_{2})^{2}+\cdots+(\xi^{\prime}_{n}-x_{n})^{2}=(a_{j}\tau^{\prime})^{2}$,
and $dS^{\prime}_{n,j}$ is its surface element. $\xi\in\mathbb{R}_{n}$ is the
integral variable on the hypersphere
$(\xi_{1}-x_{1})^{2}+(\xi_{2}-x_{2})^{2}+\cdots+(\xi_{n}-x_{n})^{2}=(a_{j}\tau)^{2}$,
and $dS_{n,j}$ is its surface element.

Proof. In Theorem 9, let $P(\partial_{x})=\Delta_{n}$, then Theorem 11 is
proved by the substitution of (93).

Similarly, we can easily obtain explicit solutions of the Cauchy problem of
more complex partial differential equations. Therefore, using the method of
abstract operators, we have established a general theory of initial value
problems for linear higher-order partial differential equations.

###  3.3 Initial-boundary value problem and Hilbert space

Clearly, similar to $\sinh(h\partial_{x})\;\mbox{and}\;\cosh(h\partial_{x})$,
the abstract operators

$\frac{\sinh\left(at\Delta_{n}{}^{1/2}\right)}{a\Delta_{n}{}^{1/2}}\quad\mbox{and}\quad\cos\left(at\Delta_{n}{}^{1/2}\right)=\frac{\partial}{\partial{t}}\,\frac{\sinh\left(at\Delta_{n}{}^{1/2}\right)}{a\Delta_{n}{}^{1/2}}$

on the left side of (93) are bounded operators in a Hilbert space, and can act
on the whole Hilbert space. Thus we can attach proper boundary conditions to
the initial value problem introduced by Theorem 7, which makes the given
functions $f(x,t),\;\varphi_{j}(x)$ in Eq (91) and (94) become functions with
boundary conditions. Then solving this initial-boundary value problem comes
down to expressing $f(x,t),\;\varphi_{j}(x)$ in a Hilbert space $H$ within the
given domain $\Omega$.

For the initial-boundary value problem, the operator $P(\partial_{x})$ must
have the characteristic function related to boundary conditions, in order to
expand the known function $f(x,\tau),\;\varphi_{r}(x)$ in (71), (80) and (82)
by using the characteristic function of $P(\partial_{x})$. $P(\partial_{x})$
in Theorem 7, Theorem 8 and Theorem 9 can be variable-coefficient partial
differential operators. For instance, if $P(\partial_{x})$ is a self-adjoint
operator defined on a Hilbert space $H$, then the abstract operators

$\frac{\sinh(tP(\partial_{x})^{1/2})}{P(\partial_{x})^{1/2}}\quad\mbox{and}\quad\cosh(tP(\partial_{x})^{1/2})=\frac{\partial}{\partial{t}}\,\frac{\sinh(tP(\partial_{x})^{1/2})}{P(\partial_{x})^{1/2}}$

are the continuous operators on the Hilbert space. In this case, we can attach
proper boundary conditions to the initial value problems in (70), (79) and
(81). Therefore, the given function $f(x,t),\varphi_{r}(x)$ becomes a function
with boundary conditions, and can be expressed in the Hilbert space $H$ within
the given domain $\Omega$. In order to solve the corresponding initial-
boundary value problem, we need to solve the characteristic value problem of
$P(\partial_{x})$ under given boundary conditions to determine a set of
orthogonal functions, which generates a linear manifold of the Hilbert space,
thus $f(x,t),\varphi_{r}(x)$ can be expressed in the Hilbert space.

In (71), (80) and (82), if $f(x,\tau),\varphi_{r}(x)\in{L^{2}}(\Omega)$, and
$P(\partial_{x})$ is a second-order linear self-adjoint elliptic operator,
namely

$P(\partial_{x})u=\sum^{n}_{i,j=1}\frac{\partial}{\partial{x}_{j}}\left(a_{ij}(x)\frac{\partial{u}}{\partial{x}_{i}}\right)+c(x)u,\quad{x}\in\Omega\subset\mathbb{R}^{n},$
(96)

then boundary conditions can be added for the definite solution problems (70),
(79) and (81): $\overline{B}u|_{\partial\Omega}=0$ representing
$u|_{\partial\Omega}=0$ or

$\left[\sum^{n}_{i,j=1}a_{ij}(x)\frac{\partial{u}}{\partial{x}_{j}}\cos\langle\mathbf{a},x_{i}\rangle+b(x)u\right]_{\partial\Omega}=0.$

Where $\mathbf{a}$ is the unit outward normal of $\partial\Omega$. Thus this
kind of initial boundary value problems boils down to solving the
characteristic value problem of the first boundary value problem of second-
order linear self-adjoint elliptic operator:

$\left\\{\begin{array}[]{l@{\qquad}l}\displaystyle\sum^{n}_{i,j=1}\frac{\partial}{\partial{x}_{j}}\left(a_{ij}(x)\frac{\partial{u}}{\partial{x}_{i}}\right)+c(x)u=-\lambda{u},\quad{x}\in\Omega\subset\mathbb{R}^{n},\\\
\displaystyle\overline{B}u|_{\partial\Omega}=0\end{array}\right.$ (97)

Reference [7] quotes the following theorem:

Let $\Omega\subset\mathbb{R}^{n}$ be a bounded open domain, and
$\partial\Omega$ be smooth. Let $a_{ij}=a_{ji}$, and there exists $\theta>0$
such that

$\sum^{n}_{i,j=1}a_{ij}(x)\xi_{i}\xi_{j}\geq\theta|\xi|^{2},\quad{x\in\Omega}.$

And let
$a_{ij}\in{C}^{1}(\overline{\Omega}),\;c(x)\in{C}(\overline{\Omega}),\;b(x)\in{C}(\partial\Omega)$,
then (97) has the following countable characteristic values:

$0\leq\lambda_{1}\leq\lambda_{2}\leq\cdots\leq\lambda_{\nu}\leq\cdots,\quad\lim_{\nu\rightarrow\infty}\lambda_{\nu}=\infty$

(If $(a_{ij})=I$ is a unit matrix, then $\lambda_{1}=0$ when $b(x)=c(x)=0$.
When $b(x)\geq 0,\,c(x)\geq 0$ and one of them does not identically equal to
zero, $\lambda_{1}>0$) and the corresponding characteristic functions
$e_{1}(x),e_{2}(x),\cdots,e_{\nu}(x),\cdots,$ satisfy

$\sum^{n}_{i,j=1}\frac{\partial}{\partial{x}_{j}}\left(a_{ij}(x)\frac{\partial{e_{i}}}{\partial{x}_{i}}\right)+c(x)e_{i}=-\lambda_{i}{e_{i}},\quad(e_{i},\,e_{j})=\delta_{ij}$
(98)

and $\\{e_{j}(x)\\}^{\infty}_{j=1}$ is complete in $L^{2}(\Omega)$, that is
for an arbitrary $f(x)\in{L^{2}}(\Omega)$, there exists $c_{j}$ such that

$\lim_{m\rightarrow\infty}\|f-\sum^{m}_{j=1}c_{j}e_{j}\|_{L^{2}(\Omega)}=0.$

Therefore, for (71), (80) and (82), by
$f(x,\tau),\varphi_{r}(x)\in{L^{2}}(\Omega)$ we also have

$\lim_{m\rightarrow\infty}\|f(x,\tau)-\sum^{m}_{j=1}c_{j}(\tau)e_{j}(x)\|_{L^{2}(\Omega)}=0.$
$\lim_{m\rightarrow\infty}\|\varphi_{r}(x)-\sum^{m}_{j=1}c_{j}e_{j}(x)\|_{L^{2}(\Omega)}=0.$

Clearly, based on (96) and (98), we have the extension of abstract operators
in Hilbert space:

$e^{\tau{a_{j}}P(\partial_{x})}e_{i}(x)=e^{-\tau{a_{j}}\lambda_{i}}e_{i}(x),\quad\frac{\sinh(\tau{a_{j}}P(\partial_{x})^{1/2})}{a_{j}P(\partial_{x})^{1/2}}e_{i}(x)=\frac{\sin(a_{j}\sqrt{\lambda_{i}}\,\tau)}{a_{j}\sqrt{\lambda_{i}}}e_{i}(x).$
(99)

Where $x\in\Omega\subset\mathbb{R}^{n},\;j=1,2,\cdots,m$. Then we have

$\lim_{m\rightarrow\infty}\|e^{\tau^{\prime}{a_{j}}P(\partial_{x})}f(x,\tau)-\sum^{m}_{j=1}(f,e_{j})e^{-\tau^{\prime}{a_{j}}\lambda_{i}}e_{j}(x)\|_{L^{2}(\Omega)}=0.$
$\lim_{m\rightarrow\infty}\|e^{\tau{a_{j}}P(\partial_{x})}\varphi_{r}(x)-\sum^{m}_{j=1}(\varphi_{r},e_{j})e^{-\tau{a_{j}}\lambda_{i}}e_{j}(x)\|_{L^{2}(\Omega)}=0.$
$\lim_{m\rightarrow\infty}\left\|\frac{\sinh(\tau^{\prime}a_{j}P(\partial_{x})^{1/2})}{a_{j}P(\partial_{x})^{1/2}}f(x,\tau)-\sum^{m}_{j=1}(f,e_{j})\frac{\sin(a_{j}\sqrt{\lambda_{i}}\,\tau^{\prime})}{a_{j}\sqrt{\lambda_{i}}}e_{j}(x)\right\|_{L^{2}(\Omega)}=0.$
$\lim_{m\rightarrow\infty}\left\|\frac{\sinh(\tau{a_{j}}P(\partial_{x})^{1/2})}{a_{j}P(\partial_{x})^{1/2}}\varphi_{r}(x)-\sum^{m}_{j=1}(\varphi_{r},e_{j})\frac{\sin(a_{j}\sqrt{\lambda_{i}}\,\tau)}{a_{j}\sqrt{\lambda_{i}}}e_{j}(x)\right\|_{L^{2}(\Omega)}=0.$

Thus when (70), (79) and (81) satisfy the boundary condition
$\overline{B}u|_{\partial\Omega}=0$, if the orthogonal function system
$\\{e_{j}(x)\\}^{\infty}_{j=1}$ is known, we have the solution of the
corresponding initial boundary value problem.

In conclusion, based on the method of abstract operators and comprehensively
applying the Laplace transform and Hilbert space theory, we establish a
general theory of initial value and initial boundary value problems of linear
partial differential equations, and illustrate broad application prospects of
the abstract operator theory, which will definitely have a wide and profound
influence on various scientific fields.

## References

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  * [2] G.Q. Bi, Applications of abstract operator in partial differential equation(ii), _Chinese Quarterly Journal of Mathematic_. 14(1999), 80-87
  * [3] G.Q. Bi, Operator methods in high order partial differential equation, _Chinese Quarterly Journal of Mathematics_. 16(2001), 88-101
  * [4] S.X. Chen, _Pseudodifferential Operators_ , (Second Edition), Higher Education Press, Beijing, China, 2006, pp. 8-25 (in Chinese)
  * [5] G.Q. Bi and Y.K. Bi, New properties of Fourier series and Riemann Zeta function, 2010. arXiv:1008.5046
  * [6] D.F. Connon, Some series and integrals involving the Riemann zeta function, binomial coefficients and the harmonic numbers, Volume I, 2007\. arXiv:0710.4022
  * [7] S. Wang, _Sobolev Space and Introduction Partial Differential Equations_ , Science Press, Beijing, China, 2009, pp. 156-157 (in Chinese)