When is a vector field the gradient of the pressure

Question

In Frank White's book Fluid mechanics problem 4.27 (8th ed.) a 2D-velocity field is given as

$$ \vec{v} = (2xy,-y^2) $$

Using Euler's equation (assuming stationary, incompressible flow and no gravitational field)

$$ \rho\vec{v}\cdot\nabla\vec{v} = -\nabla p $$

The pressure gradient is determined to be

$$ \nabla p = (2xy^2, 2y^3) $$

But if we set $\nabla p = (P,Q)$ then

$ \partial_y P \neq \partial_x Q $

and $\nabla p$ is not conservative and there cannot exist a pressure field $p$ compatible with the provide velocity field. Hence, the velocity field provide cannot be a solution to Euler's equation with the given assumptions. Is this correct (and hence to problem statement wrong)?

Answer 1

The velocity field with components $u = 2xy$ and $v = -y^2$ does satisfy $\nabla \cdot \mathbf{v} = \frac{\partial u}{\partial x} + \frac{\partial v }{\partial y} =0$ and so could represent a steady, two-dimensional and incompressible flow. The flow is not irrotational, however, and not a potential flow, since the vorticity component in the direction perpendicular to the plane of flow does not vanish:

$$ \omega_z = \frac{\partial v}{\partial x} - \frac{\partial u }{\partial y} = -2x$$

As you determined, it does not provide a solution to the steady, two-dimensional Euler equations since the pressure field would have to satisfy

$$\frac{\partial p}{\partial x} = -\rho\left(u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y}\right)= -2\rho xy^2, \quad \frac{\partial p}{\partial y} = -\rho\left(u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y}\right)= -2\rho y^3,$$

and there is no scalar field $p$ where $\nabla p$ has those components.

What this means is that there can be no steady, two-dimensional, incompressible, rotational and inviscid velocity field of this form. Physically no such flow with non-zero vorticity could evolve from rest by the imposition of a pressure field. There can be no production or diffusion of vorticity in two-dimensional, incompressible inviscid flow since all of the terms on the RHS of the vorticity equation are zero. That is

$$ \frac{D \mathbf{\omega}}{Dt} = \frac{\partial \mathbf{\omega}}{\partial t}+ \mathbf{v} \cdot \nabla \mathbf{\omega}= \omega \cdot \nabla \mathbf{v} + \nu \nabla^2\mathbf{v} = 0,$$

since in inviscid flow $\nu = 0$ and in two-dimensional flow we have $\mathbf{\omega}\cdot \nabla \mathbf{V} = \omega_z \frac{\partial \mathbf{v}}{\partial z}= 0$ (no vortex line stretching).

Also note that if the given velocity field was a steady solution of the Euler equations we would have

$$0 = \mathbf{v} \cdot \nabla \mathbf{\omega} = u\frac{\partial \omega_z}{\partial x} + v \frac{\partial \omega_z}{\partial y} = (2xy)(-2) +(-y^2)(0) = -4xy,$$

which is impossible.