Differential operator

A differential operator maps a function to an expression depending on finitely many of its derivatives. A linear differential operator has the form $P={\sum }_{|\alpha |\le m}{a}_{\alpha }(x){\partial }^{\alpha }$; its order is the largest derivative order with nonzero coefficient. The Laplacian is a second-order example, and boundary conditions specify the domain when defining an operator on a function space.

A linear differential operator is a finite linear combination of derivatives with specified coefficient functions. It obeys $L(\alpha f+\beta g)=\alpha Lf+\beta Lg$ for scalar constants $\alpha ,\beta$. Variable coefficients need not commute with differentiation: $D(af)=a^{\prime }f+a{f}^{\prime }$.

For a constant $a$ and a polynomial $P$, the product rule gives $D(e^{ax}v)=e^{ax}(D+a)v$. Iteration and linearity prove the displayed identity. In particular $(D-a{)}^m(e^{ax}v)=e^{ax}{v}^{(m)}$, which reduces a repeated-root constant-coefficient linear ordinary differential equation to repeated integration.

For distinct real or complex numbers ${\lambda }_j$ and positive integers $m_j$, the vector space spanned by $t^k{e}^{{\lambda }_{j}t}$, $0\le k<m_j$, is preserved by differentiation. In each ordered basis $(e^{{\lambda }_{j}t},t{e}^{{\lambda }_{j}t},\dots )$, its matrix has ${\lambda }_j$ on the diagonal and $1,2,\dots ,m_j-1$ just above it. This follows from the product rule. Rescaling to $t^k{e}^{{\lambda }_{j}t}/k!$ makes each block a Jordan block. In particular, differentiation is invertible on this vector space exactly when every ${\lambda }_j\ne 0$: the determinant is ${\prod }_j{\lambda }_j^{m_j}$.

The Euler differential operator differentiates in logarithmic position: if $r=\log x$ and $V(r)=v(e^r)$, then $Dv=V^{\prime }$. In particular $D(x^n)=n{x}^n$, so a polynomial $p(D)$ acts on the power law $x^n$ by multiplication by $p(n)$. This converts a Cauchy-Euler equation into a constant-coefficient linear differential equation. Repeated roots give additional logarithmic factors, as ordinary constant-coefficient equations give polynomial factors multiplying exponentials.

For real $\beta$, the displayed boundary value problem with $v(\xi )\to 0$ as $\xi \to \infty$ has a solution exactly when $\beta >0$. Put $a={\beta }^{1/3}>0$. Its unique real solution isIndeed the Euler differential operator has characteristic roots $0,a,-a/2\pm i\sqrt{3}a/2$. Decay retains only the conjugate pair, and the two endpoint conditions determine their sine combination. For negative $\beta$ only one root has negative real part, so the two boundary conditions are incompatible. At zero every solution is a cubic polynomial in $\log \xi$, excluding nonzero decaying solutions. Complex exponents thus produce an oscillatory but decaying power-law envelope.

For a commutative $k$-algebra $R$, a relative differential operator is a $k$-linear endomorphism. Set $D_k^{-1}(R)=0$, and define order at most $m$ by $[P,m_a]\in D_k^{m-1}(R)$ for every $a\in R$, where $m_a$ means multiplication by $a$. Order zero is multiplication by an element of $R$. The finite-order operators form the ring $D_k(R)$ under composition. Specifying the base is essential: absolute operators on a complex polynomial ring include derivations of the coefficient field and need not be complex-linear.

A formal pseudodifferential operator is a series in integer powers of $\partial =d/dx$, bounded above but possibly infinite towards negative powers. Its coefficients are smooth functions. Multiplication uses the formal pseudodifferential composition rule. Negative powers are formal inverses, not globally defined integration operators on all smooth functions. Every coefficient in a product is a finite sum, so multiplication is defined without analytic convergence.

Treat $k$ as a formal spectral parameter near infinity. If $P={\sum }_{j\ge 0}{g}_j{\partial }^{\alpha -j}$, the formal pseudodifferential composition rule gives $Pw=k^{\alpha +\beta }{e}^{kx}{\sum }_{s\ge 0}{k}^{-s}{\sum }_{i+j+r=s}{g}_j\left(\genfrac{}{}{0ex}{}{\alpha -j}{r}\right)w_i^{(r)}$. Each coefficient contains finitely many terms. This describes a formal module action, not a claim that the infinite series converges at finite $k$.

For scalar coefficients, leading symbols commute. If $P=a{\partial }^p+\cdots$ and $Q=b{\partial }^q+\cdots$, their product has leading coefficient $ab$, and their commutator has coefficient $pa{b}^{\prime }-qb{a}^{\prime }$ at order $p+q-1$. Thus product order is at most $p+q$, with equality when $ab$ is not identically zero; commutator order is at most $p+q-1$, with equality when its displayed coefficient is not identically zero. Constant-coefficient operators commute. Nonzero smooth functions with disjoint supports show why product orders need not add over $C^{\infty }(\mathbb{R})$.

The formal residue is the coefficient of ${\partial }^{-1}$ after coefficients have been moved to the left by the formal pseudodifferential composition rule. It is not an analytic residue at a point. For scalar coefficients, the residue of a commutator is a total derivative; integrating it with vanishing boundary terms produces the cyclic Adler trace.

Integrate the formal pseudodifferential residue over a period, or over the real line when the integral converges and the relevant boundary terms vanish. More algebraically, apply a linear functional on the coefficient algebra that annihilates derivatives. For $A=a{\partial }^p$, $B=b{\partial }^q$ and $r=p+q+1\ge 0$, the two residues are $\left(\genfrac{}{}{0ex}{}{p}{r}\right)a{b}^{(r)}$ and $\left(\genfrac{}{}{0ex}{}{q}{r}\right)b{a}^{(r)}$. Since $\left(\genfrac{}{}{0ex}{}{q}{r}\right)=(-1{)}^r(\genfrac{}{}{0ex}{}{p}{r})$, integration by parts proves $\mathrm{Tr}(AB)=\mathrm{Tr}(BA)$. For $r<0$ both residues vanish. Bounded orders ensure only finitely many monomial pairs contribute to the residue of a product.

For a fixed formal pseudodifferential operator $P={\sum }_j{p}_j{\partial }^j$ and a monic differential operator $L={\partial }^m+{\sum }_{l=0}^{m-1}{u}_l{\partial }^l$, cyclicity of the Adler trace gives $l_P(L)=\mathrm{Tr}(PL)$ with $c=\int p_{-m-1}dx$ and $a_l=\mathrm{res}({\partial }^{l}P)={\sum }_{k=0}^l\left(\genfrac{}{}{0ex}{}{l}{k}\right)p_{k-l-1}^{(k)}$. The function is affine in the variable coefficients; its variational derivative with respect to $u_l$ is $a_l$.

The generalized Leibniz rule extends composition of differential operators to all integer $n$. Define $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=n(n-1)\cdots (n-r+1)/r!$. For nonnegative $n$ the sum stops at $r=n$. For $n=-s<0$, its coefficients are $(-1{)}^r(\genfrac{}{}{0ex}{}{s+r-1}{r})$; in particular ${\partial }^{-1}f=f{\partial }^{-1}-f^{\prime }{\partial }^{-2}+f^{\prime \prime }{\partial }^{-3}-\cdots$. Applying $\partial$ on the left cancels consecutive derivative terms and recovers multiplication by $f$, which explains the inverse expansion.

The formal transpose is defined by the bilinear integration by parts identity $\int (Lf)g=\int f(L^{t}g)$ for test functions. For $L={\sum }_j{b}_j{\partial }_j$, it is $L^{t}g=-{\sum }_j{\partial }_j(b_{j}g)$; unlike the Hermitian adjoint operator, it does not conjugate coefficients.

A differential operator is in divergence form when its highest derivatives occur inside a divergence, for example $Lu=\nabla \cdot (A\nabla u)$ or $(p(x)u^{\prime }{)}^{\prime }$. This form makes integration by parts express the operator through a lower-order bilinear form and boundary flux.