Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
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
up vote 2 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.