In a paper by Girimaji (2006), the author begins by stating the instantaneous incompressible flow equations which are as follows:
$\frac{\partial V_i}{\partial t} + V_j \frac{\partial V_i}{\partial x_j} = - \frac{\partial p}{\partial x_i} + \nu \frac{\partial^2 V_i}{\partial x_j \partial x_j}$ and $\frac{\partial^2 p}{\partial x_i \partial x_i} = - \frac{\partial V_i}{\partial x_j} \frac{\partial V_j}{\partial x_i}$.
The first equation is simply the incompressible N-S equation without body force but I have not come across the second equation (Laplace of pressure on the left-hand side) before and am wondering what it is, especially the right-hand term.
Could someone identify the second equation, please? Also, what difference does 'instantaneous' make to these equations?
The equation of continuity for incompressible flow is $\nabla \cdot \mathbf{V} = 0,$ and in terms of Cartesian components using the Einstein convention (summation over repeated indices) we have
$$\frac{\partial V_i}{\partial x_i} = 0$$
In other words, the velocity field is solenoidal (divergence free).
Taking the divergence of both sides of the Navier-Stokes equation, we get
$$\frac{\partial}{\partial x_i}\left(\frac{\partial V_i}{\partial t} + V_j \frac{\partial V_i}{\partial x_j}\right) = \frac{\partial}{\partial x_i}\left(- \frac{\partial p}{\partial x_i} + \nu \frac{\partial^2 V_i}{\partial x_j \partial x_j}\right),$$
which reduces to
$$\tag{1}\frac{\partial}{\partial t} \left(\frac{\partial V_i}{\partial x_i} \right)+ \frac{\partial}{\partial x_i}\left(V_j \frac{\partial V_i}{\partial x_j} \right) = - \frac{\partial^2 p}{\partial x_i\partial x_i} + \nu \frac{\partial^2 }{\partial x_j \partial x_j}\left(\frac{\partial V_i}{\partial x_i} \right) $$
Since the divergence of the velocity is zero, the first term on the LHS of (1) and the second term on the RHS of (1) are zero. Removing these terms and rearranging we get
$$\tag{2}\frac{\partial^2 p}{\partial x_i\partial x_i} = - \frac{\partial}{\partial x_i}\left(V_j \frac{\partial V_i}{\partial x_j} \right) = - \frac{\partial V_j}{\partial x_i} \frac{\partial V_i}{\partial x_j}- V_j \frac{\partial}{\partial x_i}\left(\frac{\partial V_i}{\partial x_j} \right)\\ = - \frac{\partial V_j}{\partial x_i} \frac{\partial V_i}{\partial x_j}- V_j \frac{\partial}{\partial x_j}\left(\frac{\partial V_i}{\partial x_i} \right)$$
Substituting again with $\frac{\partial V_i}{\partial x_i} = 0$ on the RHS of (2), it follows that
$$\frac{\partial^2 p}{\partial x_i\partial x_i} = - \frac{\partial V_j}{\partial x_i} \frac{\partial V_i}{\partial x_j}$$