Definition of pressure in Euler's equation for incompressible inviscid fluid

Question

In fluid dynamics, Euler's equations describe an inviscid fluid. For an incompressible fluid with a constant and uniform density it reads (cf. Wikipedia article):

$$ \begin{align} {\partial\mathbf{u} \over \partial t} + (\mathbf{u} \cdot \nabla) \mathbf{u} &= -\frac 1 {\rho_0} \nabla p + \mathbf{g} \\ \nabla \cdot \mathbf{u} &= 0 \end{align} $$

In order to completely define the problem, e.g. to numerically simulate it, I will also need to know how $p$ is defined in terms of $\mathbf{u}$, the function I want to solve for. To my surprise, none of the places talking about Euler's equations I've found so far give a definition of $p = p(\mathbf{x}, \mathbf{u}, t)$...

Answer 1

For incompressible fluids the $p$ is what it needs to be in order to satisfy $\nabla\cdot {\bf v}=0$. In other words you do not use $p$ to solve for the motion, but instead use the motion to find $p$.

A brief discussion of this, and the strategy for solving incompressible flow, is in exercise 67 in our book. A draft version can be found here. Th extercise in on on page 228 in the draft.

Answer 2

If you like, you can eliminate pressure altogether. Start with the inviscid momentum equation

$$\frac{\partial \mathbf{u}}{\partial t}+(\mathbf{u} \cdot \nabla) \mathbf{u}=-\frac{1}{\rho_{0}} \nabla p+\mathbf{g}.$$

Now, take the divergence of both sides and note that

$$\nabla \cdot \frac{\partial \mathbf{u}}{\partial t} = \frac{\partial }{\partial t} (\nabla \cdot \mathbf{u}) = 0, \\ \nabla \cdot \mathbf{g} = 0$$

so we get

$$\frac{1}{\rho_0} \nabla^2 p = -\nabla \cdot (\mathbf{u} \cdot \nabla \mathbf{u}).$$

Inverting this expression can be tricky as it depends on your geometry. For an infinite domain, the solution is

$$p(\mathbf{x})=\frac{\rho_0}{4 \pi} \int \frac{[\nabla \cdot(\mathbf{u} \cdot \nabla \mathbf{u})]^{\prime}}{\left|\mathbf{x}-\mathbf{x}^{\prime}\right|} \mathrm{d} \mathbf{x}^{\prime}.$$

Plugging this back into the Euler equation yields an integro-differential equation of a single variable.

Answer 3

You're correct in that you do need to know $\nabla p$ in order to solve the equation. What form that takes will depend on your physical set-up.

If you only have hydrostatic pressure variation then there is only a pressure gradient in the $z$ direction, with $\frac{dp}{dz}=-\rho g$. When modelling pipe flow, the pressure gradient is typically only in the $x$ direction and is a known function that we impose.

However, it's also common that we don't know anything about the form of the pressure gradients and therefore we have to take a different approach. By taking the curl of the equations we get equations in terms of the vorticity $\omega=\nabla \times u$ instead of the velocity. As the curl of the gradient of a scalar function is identically zero, $\nabla \times (\nabla p)\equiv0$, this removes the need to know about the pressure field. Then you can solve for the vorticity field and calculate the velocities afterwards.