Canonical pressure from an integrable tensor

For a matrix field $g=(g_{ij})$ with entries in $L^1(\mathbb R^n)$ , define its canonical pressure $\pi_*$ by

\widehat{\pi_*}(\xi)=-\sum_{i,j}\frac{\xi_i\xi_j}{|\xi|^2}\widehat{g_{ij}}(\xi)\quad(\xi\ne0).

The right side is a bounded measurable function. Its inverse Fourier transform is in $H^{-s}$ for every $s>n/2$ , by the integrable-input Sobolev bound.

Pressure equation

The definition gives

-\Delta\pi_*=\sum_{i,j}\partial_i\partial_jg_{ij},\qquad \|\pi_*\|_{H^{-s}}\le C_{n,s}\sum_{i,j}\|g_{ij}\|_1.

For inputs also in $L^2$ , this agrees with $\sum R_iR_jg_{ij}$ . For an incompressible velocity, taking divergence of the momentum equation produces this pressure source with $g=u\otimes u$ , when the other terms have zero divergence. A difference of quadratic tensors gives a difference-pressure source.

Actual pressures

Another distributional solution differs by a harmonic distribution. Identifying its gradient with $\nabla\pi_*$ requires a global growth or Sobolev bound on that gradient; the Poisson equation alone does not supply it.