Identifying pressure gradients without a pressure growth assumption

Statement

On $\mathbb R^3\times(0,T)$ , suppose $w\in L^\infty_tL^2_x$ , $g\in L^\infty_tL^1_x$ , and a distribution $\pi$ satisfy

\partial_t w+\operatorname{div}g=\nu\Delta w-\nabla\pi,\qquad \nabla\cdot w=0.

Let $\pi_*$ be the canonical pressure of $g$ . Then $\nabla\pi=\nabla\pi_*$ in space-time distributions. No spatial growth assumption on $\pi$ is required.

Time-averaged global bound

For $a\in C_c^\infty(0,T)$ , integration against the time test function gives

\int a\nabla\pi\,dt =\nu\Delta\int aw\,dt+\int a'w\,dt-\operatorname{div}\int ag\,dt.

The three terms belong to $H^{-2}$ , $L^2$ , and $H^{-3}$ , respectively, using $L^1\subset H^{-2}$ in dimension three. Also $\pi_*\in L^\infty_tH^{-2}_x$ . Thus $h_a=\int a(\nabla\pi-\nabla\pi_*)\,dt$ belongs to $H^{-3}$ .

Taking divergence of the equation yields $\Delta\pi=-\sum_{i,j}\partial_i\partial_jg_{ij}=\Delta\pi_*$ , so $\Delta h_a=0$ . The harmonic Sobolev vanishing theorem gives $h_a=0$ . Testing in space and then in time, and using density of finite sums of product test functions, proves the space-time claim. A time-dependent spatial constant in pressure is still allowed because its gradient is zero.