Let us consider the Navier-Stokes equation in $\mathbb{R}^3$ subjected to no gravitational forces provided by the formula:
\begin{align} \dfrac{\partial }{\partial t} \textbf{u} + \left(\textbf{u}\cdot \nabla \right)\textbf{u} &= \nu \nabla^{2} \textbf{u}-\dfrac{1}{\rho} \nabla p \\ \nabla \cdot \textbf{u} &= 0 \\ \textbf{u}(x,0) &= \textbf{u}^0(x) \end{align}
Here $\rho>0$ is a scalar which represents the fluids density and $\nu>0$ is a scalar which denotes the fluids kinetic viscosity. If We compute the divergence of the first equation we find
\begin{equation} \nabla \cdot \biggl[\left(\textbf{u}\cdot \nabla \right)\textbf{u} - \nu \nabla^{2} \textbf{u}\biggr] = -\dfrac{1}{\rho} \nabla^2 p \end{equation}
here $\nabla^2$ is the Laplacian operator. So the diffusion of pressure is the divergence of this non-linear vector field. Whenever the Laplacian operator has an inverse, say $(\square)^{-1}$ for simplicity), one can apply this function to obtain
$$(\square)^{-1} \biggl(\nabla \cdot -\rho \biggl[\left(\textbf{u}\cdot \nabla \right)\textbf{u} - \nu \nabla^{2} \textbf{u}\biggr]\biggr) = p.$$ This is a mapping which is completely determined by the velocity. Subsituting this into the original equation one then finds
\begin{equation} \dfrac{\partial }{\partial t} \textbf{u} + \left(\textbf{u}\cdot \nabla \right)\textbf{u} = \nu \nabla^{2} \textbf{u}+ (\square)^{-1} \biggl(\nabla \cdot \biggl[\left(\textbf{u}\cdot \nabla \right)\textbf{u} - \nu \nabla^{2} \textbf{u}\biggr]\biggr) \end{equation}
This is really the inverse Laplacian applied to the divergence of non-linear term. So it is the opposite of the diffusion the divergence of linear term. Any approach to continue this or how you could study the behavior of such a velocity field in this condition