$$\texttt{D}$$
ⅅ 𝔏 𝔏 𝔏 ϕ ϵ = ϵ … theta: ϑ = ϑ
a: 𝒶 𝒶 𝒶 and alpha: 𝛂 𝛂

Need letter: $$a$$ 𝑎 𝑎
ƒ
Fourier: 𝔉ℱ ∞
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THIN SPACE:
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EQUAL TO BY DEFINITION D: ⫐ D     …
Complex plane: ℂ Real: ℝ ℝ 𝓗(H) ··· → ∞

# 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.

# Derivative Operator

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 for the differential operator due to L. Euler:     $$\texttt{D} = {\text d}/{\text d}x \quad\mbox{or} \quad\texttt{D} = {\text d}/{\text d}t ,$$ 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. The set of continuously differential functions on open interval (𝑎, b) is denoted by ℭ¹(𝑎, b) or C¹(𝑎, b); however, we prefer to utilize the former one because letter C is usually used to denote a constant. Therefore, the differential operator is a mapping:
$\texttt{D} : ℭ^1 (a,b) \to ℭ^0 (a,b) \equiv ℭ(a,b) .$
So 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:
$\texttt{D}^0 = \texttt{I}, \qquad \texttt{D}^{n+m} = \texttt{D}^{n} \texttt{D}^{m} = \texttt{D}^{m} \texttt{D}^{n} = \frac{{\text d}^{n+m}}{{\text d} x^{n+m}} = \texttt{D}^{n} \texttt{D}^{m} ,$
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 (it is not a vector space) 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 differential 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.

Recall that the derivative operator is an unbounded operator because it can map a bounded and smooth function into unbounded one; therefore, this operator is not suitable for fixed point statement. It is convenient to set a special symbol for the derivative operator, following L. Euler: $$\texttt{D} = {\text d} / {\text d} x .$$ Its inverse is studied in calculus, and we denote one particular inverse operator as

$\texttt{D}^{-1}_c y (x) = c + \int_0^x y(s)\,{\text d} s ,$
where c is an arbitrary but fixed constant, and the starting point of integration, s = 0, is chosen for simplicity (any point from the domain of y(s) will work). The union of all possible inverse operators $$\texttt{D}_c^{-1}$$ is called in calculus as «antiderivative» to emphasize that it is not an operator because it assigns to every input function inifinite many outputs. Of course, the derivative operator is defined on the space of smooth functions and its inverse acts in the space of integrable (for instance, in Riemann sense) functions. It should be noted that such defined operator $$\texttt{D}^{-1}$$ is only right inverse to the derivative operator: $$\texttt{D}\,\texttt{D}^{-1} = \texttt{I} ,$$ the identity operator. It is well known that application of $$\texttt{D}^{-1}$$ to the derivative yields
$\texttt{D}^{-1} \texttt{D}\,y = \texttt{D}^{-1} y' (x) = c + \int_0^x y'(s)\,{\text d} s = c + y(x) - y(0) .$
Therefore, the antiderivative operator becomes truly inverse in the space of functions that vanish at the specified point (in our case, at x = 0). This is not a surprise because we need a vector space of functions in order to add them and multiply by a scalar, which is exactly a vector space.

Example 1: 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.    ■

Fokas method

Consider linear constant coefficient differential equation subject to the boundary conditions of the third kind:
$$\label{EqFokas.1} y'' + p\,y' + q\,y = f(x) , \qquad y(0) - b_0 y' (0) = c_0 , \quad y(\ell ) + b_{\ell} y'(\ell ) = c_{\ell} .$$
For simplicity, assume that the characteristic equation λ² + pλ + q = 0 has two distincs real roots
$$\label{EqFokas.2} \lambda_1 = - \frac{p}{2} + \frac{1}{2} \,\sqrt{p^2 - aq} , \qquad \lambda_1 = - \frac{p}{2} - \frac{1}{2} \,\sqrt{p^2 - aq} .$$
Upon multiplication by an exponential function $$e^{\lambda x} ,$$ where λ is one of the roots of the characteristic equation, we can reduce the result into divergent form:
$$\label{EqFokas.3} \left( y'\,e^{\lambda x} \right)' + \left( p- \lambda \right) \left( y\,e^{\lambda x} \right)' -2p\,\lambda \,y\,e^{\lambda x} = f(x)\,e^{\lambda x} .$$
Integration of the equation \eqref{EqFokas.3} with respect to x from 0 to x, we get
$\left( y'\,e^{\lambda x} \right) - y' (0) + \left( p- \lambda \right) \left( y\,e^{\lambda x} \right) - \left( p- \lambda \right) y(0) - 2p\,\lambda \int_0^x y(\xi )\,e^{\lambda \xi} {\text d}\xi = \int_0^x f(\xi )\, e^{\lambda \xi} {\text d}\xi .$ We repeat the same integration, but now from x to ℓ:
$y' (\ell ) e^{\lambda\ell} - \left( y'\,e^{\lambda x} \right) - \left( p- \lambda \right) \left( y\,e^{\lambda x} \right) + \left( p- \lambda \right) y(\ell ) \,e^{\lambda\ell} -2p\,\lambda \,\int_x^{\ell} y(\xi )\,e^{\lambda \xi} {\text d}\xi = \int_x^{\ell} f(\xi )\, e^{\lambda \xi} {\text d}\xi .$

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

$L\left[ x, \texttt{D} \right] = x\,\texttt{D} \left( \frac{1}{x}\, \texttt{D} \right) = \texttt{D}^2 - \frac{1}{x}\, \texttt{D} .$
with inverse
$w(x) = L^{-1} \left[ x, \texttt{D} \right] f(x) = w(x_0 ) + \frac{x^2 - x_0^2}{2\,x_0}\, w' (x_0 ) + \int_{x_0}^x {\text d}t\,t \int_{x_0}^t \frac{{\text d}s}{s} \, f(s) .$
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). 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 2: ???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 3: ???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 4: 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).

Another second order differential operator:
$L\left[ r, \texttt{D} \right] = r^{-2} \texttt{D} \,r^2 \texttt{D} = \texttt{D}^2 + \frac{2}{r}\, \texttt{D} , \qquad \mbox{with} \quad \texttt{D} = \frac{\text d}{{\text d}r} .$
Its inverse is
$w(r) = L^{-1} \left[ r, \texttt{D} \right] f (r) = w\left( r_0 \right) + \left( \frac{1}{r_0} - \frac{1}{r} \right) r_0^2 w' \left( r_0 \right) + \int_{r_0}^r t^{-2} {\text d}t \int_{r_0}^t s^2 f(s)\,{\text d} s .$
Another example of singular operator:
$L\left[ x, \texttt{D} \right] = x^{-n} \texttt{D}^2 x^n = \texttt{D}^2 + \frac{2n}{x}\, \texttt{D} + \frac{n(n-1)}{x^2}\, \texttt{I} .$
Its inverse is ?????
$w(x) = L^{-1} \left[ x, \texttt{D} \right] f(x) = x^{-n} \int_{x_0}^x \left( x- t \right) t^n f(t) \,{\text d} t$

Example 5:

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

We consider the Lane--Emden operator:
$L\left[ x, \texttt{D} \right] = \texttt{D}^2 + \frac{2}{x}\,\texttt{D} + a^2 \texttt{I} = \frac{1}{x^2} \,\texttt{D} \left( x^2 \texttt{D} \left[ \cdot \right] \right) + a^2 \texttt{I} = \frac{1}{x\,\sin (ax)} \, \texttt{D} \,\sin^2 (x) \texttt{D} \, \frac{x}{\sin (ax)} .$
To find its inverse, we multiply both sides of L w = f by x sin (ax) and integrate
$\sin^2 (ax) \,\texttt{D}\, \frac{x}{\sin (ax)} \,w = \sin (ax_0 ) \left( \sin \left( ax_0 \right) - a\,x_0 \cos \left( ax_0 \right) \right) w (x_0 ) + x_0 \sin (ax_0 )\,w' \left( x_0 \right) + \int_{x_0}^x s\,\sin (as)\, f(s)\, {\text d} s .$
Then we divide by $$\sin^2 (ax)$$ and integrate to obtain
$\frac{x}{\sin (ax)} \,w (x) = \frac{x_0}{\sin (ax_0 )} \,w (x_0 ) + \frac{1}{a} \left[ \cos (ax_0 ) - \cot (ax) \right] \left[ \sin (ax_0 ) \left( \sin \left( ax_0 \right) - a\,x_0 \cos \left( ax_0 \right) \right) w (x_0 ) + x_0 \sin (ax_0 )\,w' \left( x_0 \right) \right] + \int_{x_0}^x \frac{{\text d} t}{\sin^2 (at)} \int_{x_0}^t s\,\sin (as)\, f(s)\, {\text d} s .$
Therefore, the inverse >Lane--Emden operator becomes
$w(x) = L^{-1} \left[ x, \texttt{D} \right] f (x) = + \frac{\sin (ax)}{x} \int_{x_0}^x \frac{{\text d} t}{\sin^2 (at)} \int_{x_0}^t s\,\sin (as)\, f(s)\, {\text d} s .$

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.

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