Weak formulation incompressible Euler equations

Question

A divergence-free vector field $v \in L_{loc}^2(\mathbb{R}^n_{x}\times \mathbb{R}_{t},\mathbb{R}^n)$ is said to be a weak solution of the incompressible Euler equation
$$\frac{\partial v}{\partial t} + div(v \otimes v) + \nabla p = 0\\
div(v) =0$$
if
$$ \int v \cdot \frac{\partial \phi}{\partial t} + (v \otimes v) \cdot \nabla \phi \: dx dt = 0\\
\int v \cdot \nabla \phi \: dx dt = 0$$
are satisfied for every divergence-free $\phi \in C_{c}^{\infty}(\mathbb{R}^n_{x}\times \mathbb{R}_{t},\mathbb{R}^n)$. To achieve this we first have to multiply by $\phi$ and then integrate in space and time. Then by integration by parts we get
$$\int \frac{\partial v}{\partial t} \cdot \phi – (v \otimes v)\cdot \nabla \phi + \nabla p \cdot \phi \: dx dt = 0 \\
\int v \cdot \nabla \phi \: dx dt = 0$$
But I don't understand how the rest transforms. As $\Omega \subset \mathbb{R}^n_{x}\times \mathbb{R}_{t}$ we can write $\Omega = \Omega_x \times I$ for some Interval $I \subset \mathbb{R}_{t}$. So I think with weak derivation we can get $$\int_{\Omega_{x}} \int_{t} \frac{\partial v}{\partial t} \cdot \phi \: dt dx = -\int_{\Omega_{x}} \int_{t} \frac{\partial \phi}{\partial t} \cdot v \: dt dx$$
I would transform the $\nabla p \cdot \phi$ term in the same way but I read somewhere that the pressure $p$ disappears because $v$ is incompressible. Could someone explain how?

Answer 1

I read that one can write the pressure $p$ in a form where it only depends on the velocity $v$. To get the form we need we take the divergence of the Euler equation $$ \frac{\partial}{\partial t} div(v) + div(\nabla v \cdot v) + div(\nabla p) = 0$$ Because $div(v)=0$ for an incompressible fluid and $div(\nabla p) = \Delta p$, it can be transformed to $$ 0= div(\nabla v) \cdot v + tr(\nabla v \cdot \nabla v) + \Delta p = \nabla(div(v)) \cdot v + tr(\nabla v \cdot \nabla v) + \Delta p $$ As $div(v)=0$ we then get $$\Delta p = - tr(\nabla v \cdot \nabla v)$$ So I assume that it is sufficient to find a weak soultion $v$ to the weak formulation above (with eliminated pressure) and to find then an associated pressure in a weak sense.