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The modern definition of a linear operator was first given by Giuseppe Peano for a particular case. However, it was Stefen Banach who defined an operator as a function whose domain is a set of functions. Stefan Banach (1892 – 1945)
was a Polish mathematician who is generally considered one of the world's most important and influential 20th-century mathematicians. He was one of the founders of modern functional analysis, and an original member of
the Lwów School of Mathematics. When Nazi German troops conquered Lvov in 1941, all institutions of higher education were closed to Poles. The majority of intelligent people (professors, musicians, actors, writers, painters, and
many others) were executed, but Banach survived the Nazi slaughter of Polish university professors. Stefan was born in Kraków (then part of the Austro-Hungarian Empire) and passed away in Lvov (Soviet Union) with lung cancer. Stefan was a heavy smoker, and his name was given to one famous probabilistic problem, as well as to many theorems. From 1910 to 1914 and since 1920,
he dwelt in Lvov (Lwów, in Polish). At the time Banach studied there, under Austrian control as it had been from
the partition of Poland in 1772. In Banach's youth Poland, in some sense, did not exist and Russia controlled much
of the country. Banach was given the surname of his mother, who was identified as Katarzyna Banach on his birth
certificate, and the first name of his father, Stefan Greczek. He never knew his mother, who vanished from the
scene after Stefan was baptized, when he was only four days old, and nothing more is known of her.
In the spring of 1916, Hugo Steinhaus (1887--1972) made a major impact on Banach's life. Stefan wrote his first
paper together with Steinhaus. Banach major work was the 1932 book, Théorie des opérations linéaires (Theory of
Linear Operations), the first monograph on the general theory of functional analysis.
Hugo Steinhaus was a Polish mathematician (of Jewish decent) whose book Mathematical Snapshots has been very influential.
Steinhaus obtained his PhD under David Hilbert.
By an operator we mean a transformation that maps a function
into another function. A linear operator L is an operator such that
\( L[af+bg] = aLf + bLg \) for any functions f, g and any constants
a, b. Since we mostly interested in linear differential operators, we need to start with the derivative
operator, which we denote by \( \texttt{D} . \) From calculus, we know that the result
of application of the derivative operator on a function is its derivative:
\[
\texttt{D} f (x) = f' (x) = \frac{{\text d}f}{{\text d}x} \qquad \mbox{or, if independent variable is $t$,} \quad
\texttt{D} y (t) = \frac{{\text d}y}{{\text d}t} = \dot{y}.
\]
We also know that the derivative operator and one of its inverses, \( \texttt{D}^{-1} = \int , \)
are both linear operators. It is easy to construct compositions of derivative operator recursively
\( \texttt{D}^{n} = \texttt{D} \left( \texttt{D}^{n-1} \right) , \quad n=1,2,\ldots ,\)
and their linear combinations:
where coefficients \( a_n , \ a_{n-1} , \ \ldots , \ a_1, \ a_0 , \) of the linear differential operator\( L\left[ x,\texttt{D} \right] \)
could be either some functions or even constants. Generally speaking, we should multiply the last coefficient
𝑎_{0} by the identity operator, \( {\bf I} = \texttt{D}^0 , \)
but lazy people usually drop it. Since multiplication by the identity
operator does not change the result, we can omit it. The leading coefficient a_{n} is assumed to be not identically zero.
In this case we say that the linear differential operator is of the order n.
Let us consider, for simplicity, the case n = 2. With a
function y = y(x) that is twice differentiable, we assign another
function, which we denote (L y)(x) (or L[y] or
simply Ly). L[y] is the linear differential operator
that acts on y(x) by the relation
where \( a_2 (x), \ a_1 (x), \ \mbox{ and } a_0 (x) \) are given functions, and \( a_2
(x) \ne 0. \) In mathematical terminology, L is an
operator that acts on
functions; that is, there is a prescribed recipe for associating with
each function y(x) a new function (L y)(x). In other words, an operator L
assigns to each function y(x) having two derivatives a new function
called (L y)(x). Therefore, the concept of an operator coincides with
the concept of a ``function of a function.''
With such defined linear differential operator, we can rewrite any linear differential equation in operator form:
If the right-hand term (also called the driving function) is not identically zero, \( f(x) \ne 0 , \)
we call the above equation nonhomogeneous
(or inhomogeneous). The equation, which is referred to as
the homogeneous equation, with an identically zero
driving term is of particular interest:
The set of all solutions of the homogeneous equation \( L\left[ x,\texttt{D} \right] y =0 \) is
called the kernel (or nullspace) of the differential operator. In
other words, any solution of the above homogeneous
linear differential equation belongs to the kernel of the corresponding differential operator.
Example 1:
Let \( L\left[ x,\texttt{D} \right] = \texttt{D} - x^2 \)
be a first order linear differential operator (with variable coefficient). Find its kernel, denoted by ker(L).
Solution: Let \( y \in \mbox{ker}(L) , \) then by the definition of the kernel,
where n is a positive integer and \(
\texttt{D} = {\text d}/{\text d}x \) is the derivative operator, that
is, \( \texttt{D} y = y' = {\text d}y/{\text d}x \quad\mbox{and}\quad \texttt{D}^2 y = y''. \) The corresponding differential equation
\[
U_n \left[ x,\texttt{D} \right] y \equiv \left( 1 - x^2 \right) y'' -x\, y' + n(n+2)\, y =0
\]
We summarize our observation in the following theorem:
Theorem 1:
The kernel of a linear differential operator of order n
with continuous coefficients in some interval
is n-dimensional and it is spanned on n
linearly independent functions
\( y_1 (x), \ y_2 (x), \ \ldots , y_n (x) . \) Then any solution of the homogeneous
linear differential equation \( L\left[ x,\texttt{D} \right] y =0 \) can be represented as
a linear combination of these functions
where the line over f(x) denotes the complex conjugate of f(x). If one moreover adds the condition that f or g vanishes for x → 𝑎 and x → b, one can also define the adjoint operator
to the n-order differential operator \eqref{EqBanach.1}. With another inner product, we will get another formular for the adjoint operator.
A (formally) self-adjoint operator is an operator equal to its own (formal) adjoint.
For a second order linear differential operator \eqref{EqBanach.2}, correspond the adjoint operator is
Here \( W[x] = W \left( x_0 \right) \exp \left\{ -\int a_1 (x)/a_2 (x)\,{\text d}x \right\} \) is the Wronskian (arbitrary multiplier is ommited) of the operator \eqref{EqBanach.2}.
⧫
\[
L\left[ x, \texttt{D} \right] = x\, \texttt{D}^2 + \left( c - x \right) \texttt{D} + a\, \texttt{I},
\qquad\mbox{its adjoint:} \qquad x\, u'' + \left( 2 - c + x \right) u' + \left( 1-a \right) u = 0 .
\]
■
Operator Method
Differential equations play a very significant role
both in mathematics and in physics because they
describe a very wide spectrum of physical phenomena.
Therefore, the construction of solutions of differential
equations presents the very significant problem. This web site gives an introduction to inverse differential operators.
It is convenient to introduce the notation \( \texttt{D} = {\text d}/{\text d}x \)
for the derivative operator. The main problem with definding its inverse is the kernel of the derivative operator which is
not zero and it is spanned on constants. From calculus it is known that
the antiderivative
\[
\texttt{D}^{-1} f (x) = \int {\text d} x\, f(x) + C ,
\]
depends on an arbitrary constant C; it is only the right inverse of the derivative operator:
\[
\texttt{D}\,\texttt{D}^{-1} f (x) = \texttt{D}\int {\text d} x\, f(x) + \texttt{D}\, C =f(x),
\]
but
\[
\texttt{D}^{-1}\texttt{D}\, f (x) = \int {\text d} x\, f'(x) + C = f(x) +C .
\]
Therefore, to determine the inverse derivative operator uniquely, one needs to restrict it on the appropriate set of functions.
For instance, we will see later in Chapter 6 that the Laplace transform provides the tool to define functions of
the derivative operator acting on the space of functions on half line \( [0, \infty ) \) so that
f(0) = 0. In this case, the inverse operator becomes
To eliminate this troublemaker, C, we need to impose an
initial condition. For example, if we consider the set of functions that vanish at x = 0, we get the inverse operator:
This function space does not contain 1 and the exponential function \( e^{-kx} , \) considered previously.
We can apply the latter inverse operator to \( e^{-kx} -1 \) to obtain
This operator is only right inverse to our second order differential operator \( \texttt{D}^2 + k^2 , \)
but it becomes inverse if we consider a set of functions that satisfy the homogeneous initial conditions
\( y(x_0 ) =0 \quad\mbox{and} \quad y' (x_0 ) =0 . \)
Using Mathematica, we define a linear differential operator:
It is convenient to introduce the notation \( \texttt{D} = {\text d}/{\text d}x \)
for the derivative operator. The main problem with definding its inverse is the kernel of the derivative operator which is
not zero and it is spanned on constants. From calculus it is known that
the antiderivative
\[
\texttt{D}^{-1} f (x) = \int {\text d} x\, f(x) + C ,
\]
depends on an arbitrary constant C; it is only the right inverse of the derivative operator:
\[
\texttt{D}\,\texttt{D}^{-1} f (x) = \texttt{D}\int {\text d} x\, f(x) + \texttt{D}\, C =f(x),
\]
but
\[
\texttt{D}^{-1}\texttt{D}\, f (x) = \int {\text d} x\, f'(x) + C = f(x) +C .
\]
Therefore, to determine the inverse derivative operator uniquely, one needs to restrict it on the appropriate set of functions.
For instance, we will see later in Chapter 6 that the Laplace transform provides the tool to define functions of
the derivative operator acting on the space of functions on half line \( [0, \infty ) \) so that
f(0) = 0. In this case, the inverse operator becomes
To eliminate this troublemaker, C, we need to impose an
initial condition. For example, if we consider the set of functions that vanish at x = 0, we get the inverse operator:
This function space does not contain 1 and the exponential function \( e^{-kx} , \) considered previously.
We can apply the latter inverse operator to \( e^{-kx} -1 \) to obtain
This operator is only right inverse to our second order differential operator \( \texttt{D}^2 + k^2 , \)
but it becomes inverse if we consider a set of functions that satisfy the homogeneous initial conditions
\( y(x_0 ) =0 \quad\mbox{and} \quad y' (x_0 ) =0 . \)
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