Editing Pseudo-differential operator (section)

===Representation of solutions to partial differential equations===

To solve the partial differential equation

:<math> P(D) \, u = f </math>

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

:<math> P(\xi) \, \hat u (\xi) = \hat f(\xi). </math>

If the symbol ''P''(&xi;) is never zero when &xi;&nbsp;&isin;&nbsp;'''R'''<sup>''n''</sup>, then it is possible to divide by ''P''(&xi;):

:<math> \hat u(\xi) = \frac{1}{P(\xi)} \hat f(\xi) </math>

By Fourier's inversion formula, a solution is

:<math>  u (x) = \frac{1}{(2 \pi)^n} \int e^{i x \xi} \frac{1}{P(\xi)} \hat f (\xi) \, d\xi.</math>

Here it is assumed that:
# ''P''(''D'') is a linear differential operator with ''constant'' coefficients,
# its symbol ''P''(&xi;) is never zero,
# both ''u'' and &fnof; have a well defined Fourier transform.
The last assumption can be weakened by using the theory of [[distribution (mathematics)|distribution]]s.
The first two assumptions can be weakened as follows.

In the last formula, write out the Fourier transform of &fnof; to obtain

:<math>  u (x) = \frac{1}{(2 \pi)^n} \iint e^{i (x-y) \xi} \frac{1}{P(\xi)} f (y) \, dy \, d\xi.</math>

This is similar to formula ({{EquationNote|1}}), except that 1/''P''(&xi;) is not a polynomial function, but a function of a more general kind.