Preface
This section is devoted to factorization technique that is essential in solving nonhomogeneous differential equations and derivation of the corresponding Green function.
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Glossary
Factorization
Occasionally, there is a second- or higher-order linear (ODE) that, by good luck or fortunate insight, can be factorized. Of course, the factorization has to be done in such a manner that it keeps the differential equation operationally intact. In differential equations, factorization means representation of a differential operator as a composition of two or more differential operators of lessier degree or just derivative operator. From the second chapter (see section on Exact equations) it is known that a first order linear operator, \( \displaystyle L\left[ x, \texttt{D} \right] = a_1 (x)\,\texttt{D} + a_0 (x) \, \texttt{I} , \) can be factored with an aid of an integrating factor:
\[
\mu \,L\left[ x, \texttt{D} \right] = \mu \left( a_1 \texttt{D} + a_0 \texttt{D}^0 \right) = \texttt{D} \left( \mu\,a_1 (x) \right) , \qquad \texttt{D} = {\text d}/{\text d}x ,
\]
where an integrator factor μ(x) can be any solution of the adjoint differential equation
\[
L^{\ast} \left[ x, \texttt{D} \right] \mu = \left[ -a_1 \texttt{D} + \left( a_0 (x) - a'_1 (x) \right) \texttt{I} \right] \mu = 0 \qquad \Longleftrightarrow \qquad - a_1 \mu' + \left( a_0 (x) - a'_1 (x) \right) \mu = 0.
\]
As usual, D0 = I is the identity operator.
In case of the second order differential operator
\begin{equation} \label{EqFactor.1}
L \left[ x, \texttt{D} \right] y(x) = a_2 (x)\, \texttt{D}^2 + a_1 (x)\, \texttt{D} + a_0 (x)\, \texttt{D}^0
\end{equation}
one may try to factor it as
\[
L \left[ x, \texttt{D} \right] = a_2 (x)\,L_1 \left[ x, \texttt{D} \right] ,L_1 \left[ x, \texttt{D} \right] = a_2 (x) \left( \texttt{D} + p(x)\, \texttt{I} \right) \left( \texttt{D} + q(x)\, \texttt{I} \right) .
\]
However, it is hard to accomplish. To show this, we consider, for simplicity,
the case when the leading coefficient is 1, so we have a normolized differential operator
\begin{equation} \label{EqFactor.2}
L_n \left[ x, \texttt{D} \right] = \texttt{D}^2 + a_1 (x)\, \texttt{D} + a_0 (x)\, \texttt{D}^0 , \qquad \texttt{D}^0 = \texttt{I} .
\end{equation}
Then we have
\[
\left( \texttt{D}^2 + a_1 (x)\, \texttt{D} + a_0 (x)\, \texttt{D}^0 \right) = \left( \texttt{D} + p(x) \right) \left( \texttt{D} + q(x) \right) .
\]
Performing multiplication and utilizing the derivative product rule, we obatin
\[
q(x) = a_1 (x) - p(x), \qquad p' (x) + a_1 \,p(x) - p(x)^2 = a_0 (x) .
\]
The latter equation is a Riccati equation in p(x), so no general formula of the solution is known.
Instead, we try to factor the second order (and then higher order) operator into a product of simple terms containing only derivatives. Note that the differential operators with variable coefficients do not commute.
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.
■
Constant coefficient differential operator
\[
L\left[ \texttt{D} \right] = a_n \texttt{D}^n + a_{n-1} \texttt{D}^{n-1} + \cdots + a_1 \texttt{D} + a_0 \texttt{D}^0 , \qquad a_n \ne 0,
\]
can always be factored; however, this factorization is not unique.
Example 2: We factorize the harmonic operator
\[
L\left[ \texttt{D} \right] = \texttt{D}^2 + \omega^2 \texttt{D}^0 = \texttt{D}^2 + \omega^2 \texttt{I} , \qquad \texttt{D}^0 = \texttt{I} ,
\]
by multiplying it by an integrating factor, which is a solution of its adjoint equation: either sinωt or cosωt. So we have
\[
\sin (\omega t)\,L\left[ \texttt{D} \right] = \frac{\text d}{{\text d}t} \left[ \sin (\omega t)\, \frac{\text d}{{\text d}t} \left( \cdot \right) - \omega\,\cos (\omega t)\,\left( \cdot \right) \right]
\]
and
\[
\cos (\omega t)\,L\left[ \texttt{D} \right] = \frac{\text d}{{\text d}t} \left[ \cos (\omega t)\, \frac{\text d}{{\text d}t} \left( \cdot \right) + \omega\,\sin (\omega t)\,\left( \cdot \right) \right]
\]
Another kind of factorization is
\[
\sin (\omega t)\,L\left[ \texttt{D} \right] = \frac{\text d}{{\text d}t} \left[ \sin^2 (\omega t)\, \frac{\text d}{{\text d}t} \frac{\left( \cdot \right)}{\sin (\omega t)} \right]
\]
and
\[
\cos (\omega t)\,L\left[ \texttt{D} \right] = \frac{\text d}{{\text d}t} \left[ \cos^2 (\omega t)\, \frac{\text d}{{\text d}t} \frac{\left( \cdot \right)}{\cos (\omega t)} \right]
\]
■
Second order differential operator
\begin{equation} \label{EqFactor.3}
\mu (x)\,L \left[ x, \texttt{D} \right] = \mu (x) \left( a_2 (x) \,\texttt{D}^2 + a_1 (x)\, \texttt{D} + a_0 (x)\, \texttt{D}^0 \right) = \texttt{D} \left( q_2 (x) \,\texttt{D}\, q_1 (x) \right) .
\end{equation}
By multiplying each side in Eq.\eqref{EqFactor.3}, we get the system of equations
\begin{align} \label{EqFactor.4}
\mu (x)\, a_2 (x) &= q_1 (x)\,q_2 (x) , \\
\mu (x)\, a_1 (x) &= 2 q'_1 (x)\,q_2 (x) + q_1 (x)\, q'_2 (x) , \label{EqFactor.5}
\\
\mu (x)\, a_0 (x) &= q'_1 (x)\, q'_2 (x) + q''_1 (x)\,q_2 (x) . \label{EqFactor.6}
\end{align}
Solving \eqref{EqFactor.4}, \eqref{EqFactor.5}, and \eqref{EqFactor.6}, for the functions q1(x), q2(x), and μ(x), it leads to
\begin{equation} \label{EqFactor.7}
q_1 (x) = \frac{1}{a_2 \mu\,W[x]} , \qquad q_2 (x) = a_2^2 \mu^2 W[x] ,
\end{equation}
where
\begin{equation} \label{EqFactor.8}
W[x] = W \left( x_0 \right) \exp \left\{ - \int \frac{a_1 (x)}{a_2 (x)}\,{\text d}x \right\}
\end{equation}
is the Wronskian of the operator \eqref{EqFactor.1} and
\( \displaystyle \mu (x) = \frac{q_1 (x)\, q_2 (x)}{a_2 (x)} \) is any solution of the homogeneous equation generated by the adjoint operator
\begin{equation} \label{EqFactor.9}
L^{\ast} \left[ x, \texttt{D} \right] \mu (x) = a_2 (x)\,\mu'' + \left( 2\,a'_2 (x) - a_1 (x) \right) \mu' + \left( a''_2 (x) - a'_1 (x) + a_0 (x) \right) \mu = 0.
\end{equation}
This leads the factorization
\begin{equation} \label{EqFactor.10}
\mu (x)\,L \left[ x, \texttt{D} \right] \left( \cdot \right) = \mu (x) \left( a_2 (x) \,\texttt{D}^2 + a_1 (x)\, \texttt{D} + a_0 (x)\, \texttt{D}^0 \right) \left( \cdot \right) = \frac{\text d}{{\text d}x} \left( a_2^2 \mu^2 W[x]\,\frac{\text d}{{\text d}x} \left( \frac{1}{a_2 \mu\,W[x]}\,\left( \cdot \right) \right) \right) .
\end{equation}
The differential operator can also be decomposed into an equivalent form containing the general linear first order multiple:
\begin{equation} \label{EqFactor.11}
\mu (x)\,L \left[ x, \texttt{D} \right] \left( \cdot \right) = \mu (x) \left( a_2 (x) \,\texttt{D}^2 + a_1 (x)\, \texttt{D} + a_0 (x)\, \texttt{D}^0 \right) \left( \cdot \right) = \frac{\text d}{{\text d}x} \left( a(x)\,\frac{\text d}{{\text d}x} \left( \cdot \right) + b(x) \,\left( \cdot \right) \right) .
\end{equation}
Then coefficients must satisfy the following system of equations
\begin{align*}
\mu\,a_2 (x) &= a(x) , \\
\mu\,a_1 (x) &= a'(x) + b(x), \\
\mu\,a_0 (x) &= b'(x) , \\
\end{align*}
and μ is any solution of the adjoint differential equation
\( L^{\ast} \left[ x, \texttt{D} \right] \mu = 0 . \) Solving the system of equations, we obtain
\[
a(x) = \mu(x)\, a_2 (x) , \qquad b(x) = \mu \left( a_1 - a'_2 \right) - \mu' a_2 .
\]
Exact second order differential equation
\begin{equation} \label{EqFactor.12}
a''_2 (x) - a'_1 (x) + a_0 (x) = 0 ,
\end{equation}
the operator \eqref{EqFactor.3} becomes exact.
Eq.\eqref{EqFactor.9} is reduced to the first order with respect to the derivative of μ, so factorization of Eq.\eqref{EqFactor.10} can be accomplished explicitely without multiplication by an integrating factor:
\begin{equation} \label{EqFactor.13}
L \left[ x, \texttt{D} \right] \left( \cdot \right) = \left( a_2 (x) \,\texttt{D}^2 + a_1 (x)\, \texttt{D} + a_0 (x)\, \texttt{D}^0 \right) \left( \cdot \right) = \frac{\text d}{{\text d}x} \left( a_2^2 W[x]\,\frac{\text d}{{\text d}x} \left( \frac{1}{a_2 \,W[x]}\,\left( \cdot \right) \right) \right) .
\end{equation}
One solution of the homogeneous equation is known
\begin{equation} \label{EqFactor.14}
L \left[ x, \texttt{D} \right] \varphi = a_2(x)\, \varphi'' + a_1 (x)\,\varphi' + a_0 (x)\,\varphi = 0
\end{equation}
is known. Then \( \displaystyle \psi (x) = \frac{\varphi (x)}{a_2 (x)\,W[x]} \) is a solution of the adjoint equation
\( \displaystyle a_2 \psi'' + \left( 2a'_2 - a_1 \right) \psi' + \left( a''_2 - a'_1 + a_0 \right) \psi = 0 . \) This leads to factorization
\begin{equation} \label{EqFactor.15}
\psi (x)\,L \left[ x, \texttt{D} \right] y = \frac{{\text d}}{{\text d}x} \left[ \frac{\varphi^2 (x)}{W[x]} \, \frac{{\text d}}{{\text d}x} \left( \frac{y}{\varphi} \right)\right] = \frac{{\text d}}{{\text d}x} \left[ a_2^2 \psi^2 (x)\,W[x] \, \frac{{\text d}}{{\text d}x} \left( \frac{y}{a_2 \psi\,W[x]} \right) \right] .
\end{equation}
For exact equation (when Eq.\eqref{EqFactor.12} is valid), its factorization is reduced to
\begin{equation} \label{EqFactor.16}
L \left[ x, \texttt{D} \right] y = \frac{{\text d}}{{\text d}x} \left[ a_0^2 W[x] \, \frac{{\text d}}{{\text d}x} \left( \frac{y}{a_0 W[x]} \right)\right] .
\end{equation}
Example 3: Consider the differential operator
\[
L \left[ x, \texttt{D} \right] = \texttt{D}^2 + x\,\texttt{D} + x\,\texttt{D}^0 , \qquad L^{\ast} \left[ x, \texttt{D} \right] = \texttt{D}^2 - x\,\texttt{D} + \left( x-1 \right) \texttt{D}^0
\]
Solving differential equation with Mathematica, we determine an integrating factor:
DSolve[u''[x] + x*u'[x] + x*u[x] == 0, u[x], x]
{{u[x] ->
E^(x - x^2/2) C[1] +
E^(-2 + x - x^2/2) Sqrt[\[Pi]/2] C[2] Erfi[(-2 + x)/Sqrt[2]]}}
DSolve[v''[x] - x*v'[x] + (x - 1)*v[x] == 0, v[x], x]
{{v[x] ->
E^x C[1] + E^(-2 + x) Sqrt[\[Pi]/2] C[2] Erfi[(-2 + x)/Sqrt[2]]}}
\[
\mu (x) = e^x .
\]
Since the Wronskian is known \( W[x] = e^{-x^2 /2} , \) we find a factorization for the given differential operator:
\[
e^x L \left[ x, \texttt{D} \right] \left( \cdot \right) = \frac{\text d}{{\text d}x} \left( e^{2x- x^2 /2} \, \frac{\text d}{{\text d}x} \left( e^{-x + x^2 /2} \left( \cdot \right) \right) \right) .
\]
■ Example 4: Kummer's equation, also known as confluent hypergeometric equation, was introduced by a German mathematician Ernst Eduard Kummer (1820--1893) in 1837.
\[
w'' + \left( b-x \right) w' - a\,w = 0 ,
\]
where 𝑎 and b are real numbers. The coefficients of the Kummer's equation satisfy condition \eqref{EqFactor.12} when 𝑎 = 1. Assuming this, we know that we don't need an integrating factor. In this case of 𝑎 = 1, Kummer's equation admits factorization
\[
\left( \texttt{D}^2 + \left( b - x \right) \texttt{D} - \texttt{I} \right) \left( \cdot \right) = \frac{\text d}{{\text d}x} \left( e^{-bx + x^2 /2} \frac{\text d}{{\text d}x} e^{bx - x^2 /2} \left( \cdot \right) \right) ,
\]
where we used the Wronskian
\[
W[x] = \exp \left\{ - \int \left( b - x \right) {\text d} x \right\} = \exp \left\{ - bx + x^2 /2 \right\} .
\]
■
Singular second order differential operator
Factorization can be applied to singular differential equations.
Consider the linear Lane–Emden equation
\[
y'' + \frac{2}{x}\,y' + a^2 y =0 \qquad \mbox{or} \qquad L \left[ x, \texttt{D} \right] y = 0,
\]
where L is the Lane--Emden operator. It can be factorized:
\[
L \left[ x, \texttt{D} \right] = \texttt{D}^2 + \frac{2}{x}\,\texttt{D} + a^2\texttt{I} = \frac{1}{x\,\sin ax} \,\frac{\text d}{{\text d}x} \left[ \sin^2 ax \, \frac{\text d}{{\text d}x} \left( \frac{x}{\sin ax} \left( \cdot \right) \right) \right] .
\]
■
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: 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.
- Bougoffa, L., On the exact solutions for initial value problems of second-orderdifferential equations, Applied Mathematics Letters, 2009, Vol. 22, pp. 1248--1251.
- Clegg, J., A new factorisation of a general second order differential equation, International Journal of Mathematical Education in Science and Technology, 2006, Vol. 37, No. 1, pp. 51--64. doi: 10.1080/00207390500186339
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