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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 $p = p(\mathbf{x}, \mathbf{u}, t)$...

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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.

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  • $\begingroup$ Thanks! Yes...what I didn't realize was that the in the system of equations stated in my question we actually have four equations in four unknowns and hence we can insert the second one into the first one (with some vector calculus voodoo) to substitute for the pressure term. I think what mostly misled me was that I was trying to find a physical interpretation for the pressure in the incompressible, inviscid case while - as far as I understand now - there isn't any, really: it is purely a mathematical definition. $\endgroup$ – Damian Birchler May 4 at 10:53
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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.

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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.

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