Preface

This section is devoted to a review of the definition of the derivative and some of its properties from the operator theory point of view.

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Glossary

Inhomogeneous Euler equations

𝔏 𝔏 𝔏 ϕ ϵ ϑ … 𝒶 𝒶 𝒶 𝛂 𝛂

Need letter: \( a \) 𝑎 𝑎 ƒ

You learn from calculus that the derivative of a smooth function f(x), defined on some interval \( (a,b) , \) is another function defined by the limit (if it exists)

\[ f'(x) \overset{\mbox{def}}{=} \frac{{\text d}f}{{\text d}x} = \lim_{h\to 0} \,\frac{f(x+h) - f(x)}{h} , \]
where we used the Lagrange and Leibniz notations for the derivative, respectively. There are known infinite many definitions of the derivative, we mention some of them:
\[ f'(x) = \lim_{h\to 0} \frac{f(x) - f(x-h)}{h} = \lim_{h\to 0} \frac{f(x+h) - f(x-h)}{2h} . \]
We will use a special notation due to L. Euler \( \texttt{D} = {\text d}/{\text d}x \quad\mbox{or} \quad\texttt{D} = {\text d}/{\text d}t \) for the derivative operator, depending on what independent variable is in use. In case of time variable t, it is a custom to utilize Newton's dot notation: \( \dot{y} = {\text d}y/{\text d}t . \) The derivative operator maps a continuously differentiable function into a continuous function. For this relation, we write \( \texttt{D} : C^1 (a,b) \to C^0 (a,b) \equiv C(a,b) \) to indicate that the derivative operator maps the set of smooth functions on some open interval into a set of continuous functions on the same interval. The derivative operator is not a bounded or continuous operator because it can transfer a bounded function into an unbounded function.

The derivative operator is a linear operator, that is,

\[ \texttt{D} \left[ u(x) + v(x) \right] = \texttt{D} \left[ u(x)\right] + \texttt{D} \left[ v(x)\right] , \qquad \texttt{D} \left[ C\,u(x)\right] = C\,\texttt{D} \left[ u(x)\right] , \]
for arbitrary constant C. We can also define integer powers of the derivative operator:
\[ \texttt{D}^0 = \texttt{I}, \qquad \texttt{D}^{n+m} = \texttt{D}^{n} \texttt{D}^{m} = \texttt{D}^{m} \texttt{D}^{n} = \frac{{\text d}^{n+m}}{{\text d} x^{n+m}} = , \]
where I is the identity operator. We would like to extend the above relation to negative indexes. So we first need to define the inverse operator, \( \texttt{D}^{-1} . \)

The inverse of the derivative is called «antiderivative» in mathematical literature mostly because it is not an operator. Why? To be an operator, \( \texttt{D}^{-1} \) should assign to every input (namely, a function) a unique output (another function). Since the kernel (or null space) of the derivative operator \( \texttt{D} \) is a one dimensional space, it assigns infinite many antiderivatives, abusing mathematicians. This follows from the identities:

\[ \texttt{D} \left[ y(x) \right] = \texttt{D} \left[ y(x) +c \right] = y' (x) \]
for any constant c. Therefore, we have infinite many inverse operators (that we mark with subscript c):
\[ \texttt{D}^{-1}_c \left[ f(x) \right] = c + \int_{x_0}^x f(s)\,{\text d}s . \]
To knock out a single inverse operator, we consider a set of continuous functions on an open interval \( (a,b) \ni x_0 \) with a prescribed condition at the point x0:
\[ C_c (a,b) = \left\{ f : f(x_0 ) = c \quad\mbox{and} \quad f \mbox{ is continuous on the interval } (a,b) \,\right\} . \] Then the derivative operator considered as a mapping
\[ \texttt{D} : C^1 (a,b) \to C_c (a,b) \subset C(a,b) \]
will be denoted by L. The latter has a unique inverse (depending on the prescribed value c): \( \texttt{D}_c^{-1} : C_c (a,b) \to C^1 (a,b) . \)

 

The operator \( \texttt{D}_c^{-1} \) is not a truly inverse, but only right inverse:
\[ \begin{split} \texttt{D}\,\texttt{D}_c^{-1} &= I \qquad \mbox{the identity operator}, \\ \texttt{D}_c^{-1} \texttt{D} &\ne I , \end{split} \]
because
\[ \texttt{D}_c^{-1} \texttt{D} f(x) = \texttt{D}_c^{-1} \, f' (x) = c + \int_{x_0}^x f' (s)\,{\text d}s = c + f(x) - f\left( x_0 \right) \ne f(x) \]
unless f(x0) = c.

Example: Let us take a constant function y(x) ≡ 1, then

\[ \texttt{D}^{-1} y (x) = \texttt{D}^{-1} \,1 = C + x , \]
where C is a constant of integration. The derivative of y(x) is identically zero, so
\[ \texttt{D}^{-1} y' (x) = \texttt{D}^{-1} \texttt{D}\,1 = \texttt{D}^{-1} 0 = C \ne 1 = \texttt{D} \, \texttt{D}^{-1} 1 = \texttt{D} \,(C+x ) . \]
So we see that \( \texttt{D} \) and \( \texttt{D}^{-1} \) do not commute (unless a condition on the constant is imposed). This means that regular indefinite integral with arbitrary constant as initial condition is only left inverse of the derivative operator.    ■

 

First order differential operator

For continuous function r(x) ∈ C and continuously differentiable function p(x) ∈ C¹, consider a first order differential operator
\[ L\left[ x, \texttt{D} \right] = r(x)\,\texttt{D}\,p(x) = r(x)\,p(x)\,\texttt{D} + r(x)\,p' (x)\, \texttt{I} = r(x) \left[ p(x)\,\texttt{D} + p' (x)\, \texttt{I} \right] , \]
where \( \texttt{I} \) is the identity operator, \( \texttt{D} \overset{\mbox{def}}{=} {\text d}/{\text d}x \) is the derivative operator, and prime denotes the derivative in Lagrange's notation. Assuming that r(x)p(x) ≠ 0, we find its inverse that depends on an arbitrary constant c:
\[ w(x) = L^{-1} \left[ x, \texttt{D} \right] f(x) = \frac{c}{p(x)} + \frac{1}{p(x)} \int_{x_0}^x \frac{f(s)}{r(s)}\,{\text d}s . \]
If the product of two functions is equal to zero at the initial point: \( r\left( x_0 \right) p\left( x_0 \right) = 0 , \) then we have a singular differential operator.

There are some examples of first order differential operators and their inverses.

\[ L\left[ x, \texttt{D} \right] = e^{-x^2} \texttt{D} e^{x^2} = \texttt{D} + 2x\,\texttt{I} , \]
with the inverse
\[ w(x) = L^{-1} \left[ x, \texttt{D} \right] f(x) = c\,e^{-x^2} + e^{-x^2} \int_0^x e^{s^2} f(s)\,{\text d}s , \]
where c is some constant. Another similar differential operator is
\[ L\left[ x, \texttt{D} \right] = e^{-x^3} \texttt{D} e^{x^3} = \texttt{D} + 3x^2\,\texttt{I} . \]
Its inverse is
\[ w(x) = L^{-1} \left[ x, \texttt{D} \right] f(x) = c\,e^{-x^3} + e^{-x^3} \int_0^x e^{s^3} f(s)\,{\text d}s . \]

 

Singular first order differential operator

If one of the functions r(x) or p(x) (or both) has a zero withing the given interval, then the linear differential operator \( L\left[ x, \texttt{D} \right] w(x) = r(x) \,\texttt{D} \left[ p(x)\,\texttt{D} w \right] \) becomes a singular one. We present some typical examples of such operators and their inverses.

Let

\[ L\left[ x, \texttt{D} \right] = \frac{1}{\tan x}\,\texttt{D}\,\tan x = \texttt{D} + \frac{1}{\sinh x\,\cosh x}\, \texttt{I} , \]
where \( \texttt{I} \) is the identity operator. Its inverse is
\[ w(x) = L^{-1} \left[ x, \texttt{D} \right] f(x) = \frac{c}{\tan x} + \frac{1}{\tan x} \int_{x_0}^x \tan s\,f(s)\,{\text d}s . \]
Another typical singular operator is
\[ L\left[ x, \texttt{D} \right] = x^{1-\gamma} \texttt{D} x^{\gamma} = x\texttt{D} + \gamma \texttt{I} \qquad \Longrightarrow \qquad w(x) = L^{-1} \left[ x, \texttt{D} \right] f(x) = c\,x^{-\gamma} + x^{-\gamma} \int_{x_0}^x s^{\gamma -1} f(s)\,{\text d}s , \]
where x0 ≠ 0.

 

Second order differential operator

Consider the general second order differential operator
\[ L\left[ x, \texttt{D} \right] y(x) = \texttt{D} \left[ p(x)\,\texttt{D} y(x) \right] + q(x)\,\texttt{D} \,y(x) + r(x)\, y = \frac{\text d}{{\text d}x} \left[ p(x)\,\frac{{\text d}y}{{\text d}x} \right] + q(x)\, \frac{{\text d}y}{{\text d}x} + r(x)\,y(x) , \]
where p(x) ∈ C¹ and other known functions q(x) and r(x) are assumed to be continuous on some fixed interval. To find the inverse operator, we multiply both sides of the equation \( L\left[ x, \texttt{D} \right] y(x) = w(x) \) by integrating factor
\[ \mu (x) = \exp \left\{ \int \frac{r(x)}{q(x)}\,{\text d}x \right\} . \]
This yields
\[ \mu (x) \,\frac{\text d}{{\text d}x} \left[ p(x)\,\frac{{\text d}y}{{\text d}x} \right] + \mu (x)\, q(x)\, \frac{{\text d}y}{{\text d}x} + \mu (x)\, r(x)\, y = \mu (x)\, w(x) . \]
Taking into accound the relation \( \mu' \, q = \mu\, r , \) we obtain
\[ \mu (x) \,\frac{\text d}{{\text d}x} \left[ p(x)\,\frac{{\text d}y}{{\text d}x} \right] + q(x) \, \frac{\text d}{{\text d}x} \left[ \mu (x)\, y \right] = \mu (x)\, w (x) , \]
so that
\[ \frac{\text d}{{\text d}x} \left[ p(x)\,\frac{{\text d}y}{{\text d}x} \right] + \frac{q(x)}{\mu (x)} \, \frac{\text d}{{\text d}x} \left[ \mu (x)\, y \right] = w(x) . \]
Suppose that the initial conditions are specified at some point x0, where the integrating factor μ(x) and its first derivative have bounded values. Upon introducing a new dependent variable z = μ y, we get
\[ \frac{\text d}{{\text d}x} \left[ p(x) \left( \frac{1}{\mu (x)} \right)' z + \frac{p(x)}{\mu (x)}\,\frac{{\text d}z}{{\text d}x} \right] + \frac{q(x)}{\mu (x)} \,\frac{{\text d}z}{{\text d}x} = w(x) . \]
If it is possible to make the coefficient of z to be a constant:
\[ p(x) \left( \frac{1}{\mu (x)} \right)' = c, \]
then our differential equation becomes
\[ \frac{\text d}{{\text d}x} \left[ s(x) \,\frac{{\text d}z}{{\text d}x} \right] + t(x)\, \frac{{\text d}z}{{\text d}x} = w(x) , \]
where
\[ s(x) = \frac{p(x)}{\mu (x)} \qquad\mbox{and}\qquad t(x) = c + \frac{q(x)}{\mu (x)} . \]
Now we multiply the above equation by another integrating factor
\[ \xi (x) = \exp \left\{ \int \frac{t(x)}{s(x)} \right\} \,{\text d}x, \]
to obtain the exact equation
\[ \frac{\text d}{{\text d}x} \left[ \xi (x)\,s(x) \,\frac{{\text d}z}{{\text d}x} \right] = \xi (x)\, w(x) . \]
Integrating with respect to x, we get
\[ \xi (x)\,s(x) \,\frac{{\text d}z}{{\text d}x} = C_1 + \int_{x_0}^x \xi (\tau )\, w(\tau )\,{\text d}\tau . \]
Second integration yields
\[ z(x) = C_2 + C_1 \int_{x_0}^x \frac{1}{\xi (x)\,s(x)}\,{\text d}x + \int_{x_0}^x \frac{1}{\xi (x)\,s(x)}\,{\text d}x \int_{x_0}^x \xi (\tau )\, w(\tau )\,{\text d}\tau . \]

 

Example: ???To be modified Consider the initial value problem

\[ x^2 \frac{{\text d}^2 y}{{\text d}x^2} + ax\,\frac{{\text d} y}{{\text d}x} + by = 3\,x^2 , \qquad y(1) , \quad y' (1) = 2. \]
We have
\[ \mu = x^{-1} , \quad \xi = x^2 , \quad h= x^3 , \quad t(x) = 2, \quad s(x) = x. \]
So z = x and y = x².    ■

 

Example: ???To be modified Consider the initial value problem

\[ \frac{{\text d}^2 y}{{\text d}x^2} - \frac{2x}{1- x^2} \, \frac{{\text d} y}{{\text d}x} + \frac{2}{1- x^2} \, y = \frac{2}{1- x^2} , \qquad y(-1) = 2 , \quad y' (-1) = -1. \]
We have
\[ \mu = x^{-1}, \quad \xi = x(1-x^2 ) , \quad h= x^2 \left( 1 - x^2 \right) , \quad t(x) = 1 - \frac{2x}{1-x^2} , \quad s(x) = x. \]
Then y = 1 - x with z = 1/x -1.    ■

 

Example: To be modified Consider the differential operator

\[ L\left[ x, \texttt{D} \right] = \texttt{D}^2 + \frac{b-x}{x}\, \texttt{D} - \frac{b}{x}\, \texttt{I} \]
Here we have p(x) = 1, q(x) = (b - x)/x, and r(x) = -b/x. Then
\[ \mu (x) = \frac{1}{b-x} , \quad \xi (x) = \frac{x^b}{b-x}\, e^{-x} , \quad h(x) = \xi (x)\, s(x) = x^b e^{-x} , \quad s(x) = b-x , \quad t(x) = \frac{(b-x)^2}{x} + 1 . \]
Then the exact solution of L y = 0 is
\[ y = \frac{1}{\mu}\,z = \frac{b-x}{b} . \]
   ■
Consider now the self-adjoint second order differential operator
\[ L\left[ x, \texttt{D} \right] y(x) = \texttt{D} \left[ p(x)\,\texttt{D} y(x) \right] - q(x)\,y(x) = \frac{\text d}{{\text d}x} \left[ p(x)\,\frac{{\text d}y}{{\text d}x} \right] - q(x)\,y(x) , \]
where p(x) and q(x) are given smooth positive functions, and \( \texttt{D}y = {\text d} y/{\text d}x \) is the derivative in the Leibniz notation. To find its inverse, we need to solve the differential equation
\[ L\left[ x, \texttt{D} \right] w(x) = f(x) . \]
For uniqueness of this operation, we have to impose two auxiliary conditions; for instance, we can take the initial conditions at some point and consider the set of continuously differentiable functions with prescribed value and its derivative:
\[ w(x_0 )= c_0 , \qquad w' (x_0 ) = c_1 . \]
Then integration gives
\[ p(x)\,\frac{{\text d}w}{{\text d}x} = p(x_0 )\,w' (x_0 ) + \int_{x_0}^x f(s)\,{\text d}s . \]
Assuming that p(x) > 0, we divide by this function and again integrate
\[ w(x) = L^{-1}\left[ x, \texttt{D} \right] f(x) = w(x_0 ) + \int_{x_0}^x \frac{{\text d}t}{p(t)} \left[ p(x_0 )\,w' (x_0 ) + \int_{x_0}^t f(s)\,{\text d}s \right] . \]
We can slightly modify the differential operator and consider
\[ L\left[ x, \texttt{D} \right] w(x) = r(x) \,\texttt{D} \left[ p(x)\,\texttt{D} w \right] = r(x)\,\frac{\text d}{{\text d}x} \left[ p(x)\,\frac{{\text d}w}{{\text d}x} \right] . \]
Assuming r(x) > 0, we make first integration
\[ p(x)\,\frac{{\text d}w}{{\text d}x} = p(x_0 )\,w' (x_0 ) + \int_{x_0}^x \frac{{\text d}t}{r(t)} \, f(t) . \]
Next integration yields the inverse operator
\[ L^{-1} \left[ x, \texttt{D} \right] f(x) = w(x_0 ) + p(x_0 )\,w' (x_0 ) \int_{x_0}^x \frac{{\text d}s}{p(s)} + \int_{x_0}^x \frac{{\text d}s}{p(s)} \int_{x_0}^s \frac{{\text d}t}{r(t)} \, f(t) \]

 

Singular second order differential operator

If one of the functions r(x) or p(x) (or both) has a zero withing the given interval, then the linear differential operator \( L\left[ x, \texttt{D} \right] w(x) = r(x) \,\texttt{D} \left[ p(x)\,\texttt{D} w \right] \) becomes a singular one. We present some typical examples of such operators and their inverses.

Let

\[ L\left[ x, \texttt{D} \right] = x\,\texttt{D} \left( \frac{1}{x}\, \texttt{D} \right) = \texttt{D}^2 - \frac{1}{x}\, \texttt{D} . \]
It is easy to verify that the above differential operator is linear:
\[ L\left[ c_1 y_1 (x) + c_2 y_2 (x) \right] = c_1 L\left[ y_1 (x) \right] + c_2 L\left[ y_2 (x) \right] , \]
for any constants c1, c2 and any two smooth functions y1(x), y2(x).

 

 

Example:

Ernst Kummer
Consider the differential equation
\[ x\, y'' + \left( \gamma -x \right) y' - \alpha\,y =0 , \]
known as confluent hypergeometric equation or Kummer's equation, introduced by Ernst Kummer in 1837. We introduce the differential operator
\[ L\left[ x, \texttt{D} \right] = x^{1-\gamma} \,\texttt{D} \left( x^{\gamma} \,\texttt{D} \right) \]
   ■

Example: because the solutions to many mathematical physics problems are expressed by the hypergeometric functions, we consider the corresponding differential equation    ■

 

We consider the nonlinear differential operator of the second order:
\[ L\left[ r, \texttt{D} \right] w(r) = \frac{1}{r} \,\texttt{D} \left[ r \left( - \texttt{D} w\right)^n \right] = \frac{1}{r} \,\frac{\text d}{{\text d}r} \left[ r \left( - \frac{{\text d}w}{{\text d}r} \right)^n \right] = -\frac{1}{r} \left( - w' \right)^{n-1} \left( w' + nr\,w'' \right) , \]
where we use the Lagrange notation for the derivative: \( w' = {\text d} w/{\text d}r . \)
Simplify[1/r*D[(r*(-D[w[r], r])^n), r]]
-(((-Derivative[1][w][r])^(-1 + n) (Derivative[1][w][r] + n r (w^\[Prime]\[Prime])[r]))/r)
Since the given differential operator is of the second order, the inverse operator will depend on two arbitrary constants; therefore, we need to consider L on a space of functions that have a specific values at a particular point and its derivative. In order to find its inverse, we need to solve the differential equation
\[ L\left[ r, \texttt{D} \right] w(r) = f(r) \qquad \Longrightarrow \qquad \frac{\text d}{{\text d}r} \left[ r \left( - \frac{{\text d}w}{{\text d}r} \right)^n \right] = r\,f(r) . \]
Integrating both sides, we get
\[ r \left( - \frac{{\text d}w}{{\text d}r} \right)^n - \left. r \left( - \frac{{\text d}w}{{\text d}r} \right)^n \right\vert_{r=r_0} = \int_{r_0}^r s\, f(s)\,{\text d}s . \]
Dividing by r and taking the n-th root, we have
\[ - \frac{{\text d}w}{{\text d}r} = \left\{ \frac{r_0}{r} \left( - w' (r_0 ) \right)^n + \frac{1}{r}\,\int_{r_0}^r s\, f(s)\,{\text d}s \right\}^{1/n} \]
Second integration yields the inverse operator
\[ L^{-1}\left[ r, \texttt{D} \right] f(r) = - \int_h^r {\text d}r \left\{ \frac{r_0}{r} \left( - w' (r_0 ) \right)^n + \frac{1}{r}\,\int_{r_0}^r s\, f(s)\,{\text d}s \right\}^{1/n} - w(h) . \]
In the above formula, you can choose r0 to be zero because the inner integral is not singular, but h should be always positive.

 

  1. Bougoffa, L., On the exact solutions for initial value problems of second-orderdifferential equations, Applied Mathematics Letters, 2009, Vol. 22, pp. 1248--1251.

 

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