Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 8 / 2

First let $n\ge 0$. The ordinary Leibniz rule, applied to a test function $h$, givesEquivalently, the composition of differential operators isFor completeness, this formula follows by induction: applying $\partial$ differentiates either the coefficient or the remaining derivative of $h$, and the two contributions combine by $\left(\genfrac{}{}{0ex}{}{n}{r}\right)+\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{r}\right)$. The initial case is multiplication by $f$.

For all integers $n$, define $\left(\genfrac{}{}{0ex}{}{n}{0}\right)=1$ and $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=n(n-1)\cdots (n-r+1)/r!$. The formal pseudodifferential composition rule isFor $n=-s<0$, $\left(\genfrac{}{}{0ex}{}{-s}{r}\right)=(-1{)}^r(\genfrac{}{}{0ex}{}{s+r-1}{r})$, so this is usually an infinite series. In particularApplying $\partial$ on the left gives $m_f$: all later coefficient derivatives cancel in pairs. This uniquely determines the normal-ordered inverse expansion. Repeating the inverse construction, or using the same binomial recurrence downward in $n$, gives every negative power. More explicitly, applying $\partial$ to the proposed $n$-series yields coefficient $\left(\genfrac{}{}{0ex}{}{n}{r}\right)+\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{r}\right)$ at $f^{(r)}{\partial }^{n+1-r}$, so it is consistent with the already determined $(n+1)$-series. Negative powers here are formal; differentiation on all of $C^{\infty }(\mathbb{R})$ has a nontrivial kernel and has no genuine two-sided inverse there.

Use the formal pseudodifferential composition rule on each coefficient $w_i(x)$ and then the exponential eigenfunction convention. For every integer $n$,ConsequentlyCollecting all terms of a fixed power of the spectral parameter gives the useful answerThe sum defining each $v_s$ is finite. For example,andThis is the action on a formal pseudodifferential wave function. It is a formal expansion near $k=\infty$, with $k^{\beta }$ carried as a formal prefactor. It does not assert convergence for finite $k$, and negative powers are not evaluated at $k=0$.

Write the two formal pseudodifferential operators in normal order aswhere $p={\alpha }_1$ and $q={\alpha }_2$. The formal pseudodifferential composition rule givesThus $\boxed{\mathrm{ord}(P_1{P}_2)=p+q}$ whenever $ab$ is not identically zero, in particular when both leading coefficients are nowhere zero. In general the rigorous statement over the specified coefficient algebra is $\mathrm{ord}(P_1{P}_2)\le p+q$.

The leading scalar coefficients commute. Subtracting the reversed composition also cancels the terms involving $a_1,b_1$, leavingTherefore the order filtration of formal pseudodifferential operators giveswith equality precisely when the displayed coefficient is not identically zero. The bound can be strict: constant-coefficient powers of $\partial$ commute, giving the zero operator, whose order can be assigned $-\infty$.

There is also a genuine qualification to product-order addition in $C^{\infty }(\mathbb{R})$. Choose nonzero smooth functions $a,b$ with disjoint compact supports and take $P_1=m_a,P_2=m_b$. Each has order zero, but their product is zero. Nonzero coefficients as elements of this algebra need not have a nonzero product. Thus an unqualified assertion of exact product order requires an additional leading-symbol hypothesis.