Pseudo-differential operator

Consider a linear differential operator with constant coefficients,

${\displaystyle P(D):=\sum _{\alpha }a_{\alpha }\,D^{\alpha }}$

which acts on smooth functions ${\displaystyle u}$ with compact support in Rⁿ. This operator can be written as a composition of a Fourier transform, a simple multiplication by the polynomial function (called the symbol)

${\displaystyle P(\xi )=\sum _{\alpha }a_{\alpha }\,\xi ^{\alpha },}$

and an inverse Fourier transform, in the form:

${\displaystyle \quad P(D)u(x)={\frac {1}{(2\pi )^{n}}}\int _{\mathbb {R} ^{n}}\int _{\mathbb {R} ^{n}}e^{i(x-y)\xi }P(\xi )u(y)\,dy\,d\xi }$

(1)

Here, ${\displaystyle \alpha =(\alpha _{1},\ldots ,\alpha _{n})}$ is a multi-index, ${\displaystyle a_{\alpha }}$ are complex numbers, and

${\displaystyle D^{\alpha }=(-i\partial _{1})^{\alpha _{1}}\cdots (-i\partial _{n})^{\alpha _{n}}}$

is an iterated partial derivative, where ∂_j means differentiation with respect to the j-th variable. We introduce the constants ${\displaystyle -i}$ to facilitate the calculation of Fourier transforms.

Derivation of formula (1)

The Fourier transform of a smooth function u, compactly supported in Rⁿ, is

${\displaystyle {\hat {u}}(\xi ):=\int e^{-iy\xi }u(y)\,dy}$

and Fourier's inversion formula gives

${\displaystyle u(x)={\frac {1}{(2\pi )^{n}}}\int e^{ix\xi }{\hat {u}}(\xi )d\xi ={\frac {1}{(2\pi )^{n}}}\iint e^{i(x-y)\xi }u(y)\,dy\,d\xi }$

By applying P(D) to this representation of u and using

${\displaystyle P(D_{x})\,e^{i(x-y)\xi }=e^{i(x-y)\xi }\,P(\xi )}$

one obtains formula (1).

To solve the partial differential equation

${\displaystyle P(D)\,u=f}$

we (formally) apply the Fourier transform on both sides and obtain the algebraic equation

${\displaystyle P(\xi )\,{\hat {u}}(\xi )={\hat {f}}(\xi ).}$

If the symbol P(ξ) is never zero when ξ ∈ Rⁿ, then it is possible to divide by P(ξ):

${\displaystyle {\hat {u}}(\xi )={\frac {1}{P(\xi )}}{\hat {f}}(\xi )}$

By Fourier's inversion formula, a solution is

${\displaystyle u(x)={\frac {1}{(2\pi )^{n}}}\int e^{ix\xi }{\frac {1}{P(\xi )}}{\hat {f}}(\xi )\,d\xi .}$

Here it is assumed that:

1. P(D) is a linear differential operator with constant coefficients,
2. its symbol P(ξ) is never zero,
3. both u and ƒ have a well defined Fourier transform.

The last assumption can be weakened by using the theory of distributions. The first two assumptions can be weakened as follows.

In the last formula, write out the Fourier transform of ƒ to obtain

${\displaystyle u(x)={\frac {1}{(2\pi )^{n}}}\iint e^{i(x-y)\xi }{\frac {1}{P(\xi )}}f(y)\,dy\,d\xi .}$

This is similar to formula (1), except that 1/P(ξ) is not a polynomial function, but a function of a more general kind.