# Idea behind factoring an operator?

Suppose if we have an operator $\partial_t^2-\partial_x^2$ what does it mean to factorise this operator to write it as $(\partial_t-\partial_x) (\partial_t+\partial x)$ When does it actually make sense and why ?

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This make sense for functions of class $C^2$, since $\partial_{tx}=\partial_{xt}$. – Davide Giraudo Jul 7 '12 at 12:18
@DavideGiraudo : Thanks , it means that $C^2$ is commutative ? isn't it ? – Theorem Jul 7 '12 at 12:22
What do you mean by "$C^2$ is commutative"? – Davide Giraudo Jul 7 '12 at 12:25
@DavideGiraudo : I was just thinking if "$C^2$"operators are commutative ? if at all it makes sense ? i might be asking a stupid question :P – Theorem Jul 7 '12 at 12:31

In the abstract sense, the decomposition $x^2-y^2=(x+y)(x-y)$ is true in any ring where $x$ and $y$ commute (in fact, if and only if they commute).

For sufficiently nice (smooth) functions, differentiation is commutative, that is, the result of derivation depends on the degrees of derivation and not the order in which we apply them, so the differential operators on a set of smooth ($C^\infty$) functions (or abstract power series or other such objects) form a commutative ring with composition, so the operation makes perfect sense in that case in a quite general way.

However, we only need $\partial_x\partial_t=\partial_t\partial_x$, and for that we only need $C^2$. Of course, differential operators on $C^2$ do not form a ring (since higher order derivations may not make sense), but the substitution still is correct for the same reasons. You can look at the differentiations as polynomials of degree 2 with variables $\partial_\textrm{something}$.

For some less smooth functions it might not make sense.

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