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It is said that we can solve differential equ $\ddot x + x = 0$ by writing it as $(d/dt + i) (d/dt - i)x=0$. Why can we do this? Certainly we cannot simply say $\frac{d^2}{dt^2} = \left(\frac{d}{dt}\right)^2$, since the linear operators, such as $d/dt$, are not the same as numbers. For example, the former is usually not commutative (though in this special case it is), and even possibly non-associative.

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Perhaps it is because that (1) $\frac{d^2}{dt^2} = \frac{d}{dt} \frac{d}{dt}$ according to the def of 2nd-order differentiation (2) the operators $d/dt$, though non-commutative, obey distribution law, similar to the matrix?

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Anyway despite the confusion about laws that linear operators follow, if I try to expand the expression $(d/dt + i) (d/dt - i)x$ from right to left then it makes sense:

$(d/dt + i) (d/dt - i)x = (d/dt + i) (dx/dt - ix)= d/dt (dx/dt - ix) + i (dx/dt - ix) \\ = \ddot x + x$

And so we can solve the ode from left to right:

From $(d/dt + i)f(t) = 0$ we get $f(t)$, then from $(d/dt - i)x(t) = f(t)$ we get $x(t)$.

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More generally speaking,

  1. are there some rules that the addition and multiplication of (partial) differentiation operators ($d/dt$ and $\partial/\partial x$, etc.) and numbers would follow?
  2. it seems that (partial) differentiation operators and matrices, both of which can be regarded as linear operators, share similarities. But they are still seemingly different, e.g. $d/dt$ and $\partial/\partial x$ may not obey associative laws while matrices do. Is there an article or book discussing the similarities and dissimilarities between the two?
  3. is there any chapter or essay that discusses in general the way of solving a higher order ode by factorizing the operators?
Charlie Chang
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1 Answers1

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You have basically answered yourself. You can consider the set $\mathscr L(X)$ of linear operators over a complex vector space $X$. And $\mathscr L(X)$ becomes an algebra when you consider pointwise addition and composition as the operations. So, as you say, $L^2$ means $L\circ L$.

The computations you made are all valid in $\mathscr L(X)$, where $X$ will be $C^\infty[0,1]$ or another suitable space depending on the initial conditions of your equation.

The relationship with matrices is not casual, as they are usually seen as linear operators and that's how their product is defined.

Martin Argerami
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  • I see (a) differentiations, integrations etc. and (b) matrices are both linear operators, and each of both can form vector spaces $\mathscr L(X)$ with multiplication defined as composition. So the algebraic structures here are similar to rings. $\quad$ The differences are that $X$ for (a) is a set of (vectors of) smooth functions, while $X$ for (b) is a set of vectors of numbers or functions. $\quad$ This implies that for $X$ being a set of fixed-length vectors of smooth functions, $\mathscr L(X)$ can contain both (a) and (b). An example might be the set of parametrized n-dim curves. – Charlie Chang Oct 12 '21 at 10:39
  • My further questions are $\quad$
    1. whether $\mathscr L(X)$ for (a) follows association laws.$\quad$
    2. whether $\mathscr L(X)$ for (a) can be a field, if we treat differentiations as inverses of integrations?
     – Charlie Chang Oct 12 '21 at 10:40
  • Not sure what you mean by "association laws". For any vector space, $\mathscr L(X)$ is an algebra (that is, a ring which is also a vector space and multiplication by scalars behaves nicely with respect to multiplication). The only way it can be a field is that $X$ is one-dimensional; otherwise, it is not commutative and it always has non-invertible elements. This can already be easily seen at the level of matrices (i.e., when $\dim X<\infty$). – Martin Argerami Oct 12 '21 at 14:21
  • Thanks. I see only when $X$ is not a set of vectors of functions $\mathscr L(X)$ can be a field. I would check the definition of algebra. Besides, it should be 'associative law'; I misspelt it. – Charlie Chang Oct 12 '21 at 17:08
  • For $\mathscr L(X)$ to exist you need $X$ to be a vector space, because linearity is defined in terms of vector-space operations. As soon as $X$ is not one-dimensional, $L(X)$ is non-commutative and it has divisors of zero, so it is definitely not a field. As for associative, composition is always associative. – Martin Argerami Oct 12 '21 at 21:30