Here I state Lamb's version of the Euler's equation for an elastic fluid:
$$ \partial_t \vec v + \vec \Omega \times \vec v + \vec \nabla \left( \frac{\|\vec v\|^2}{2} + \frac{p}{\rho} + \psi + \phi_b\right) = 0$$
where $\vec\Omega$ is the vorticity of the fluid, $p$ the pressure, $\rho$ the density, $\vec b$ is a potential body force such that $\vec b = \vec \nabla \phi_b$ and $\psi$ is Helmholt's density of energy that in our course has been defined as the fuction $\psi = \psi(\rho)$ such that $$p(\rho) = \rho^2 \frac{\partial \psi}{\partial \rho}$$
Using Lamb's equation we stated this version of Bernoulli's theorem:
Theorem: Consider an elastic eulerian fluid, then in steady flow the quantity $\frac{\|\vec v\|^2}{2} + \frac{p}{\rho} + \psi + \phi_b$ is constant along the flow lines. (Usually I've been told that the quantity $w :=\frac{p}{\rho} + \psi$ is called Gibbs' free energy)
Proof: It sufficies to show that $\vec v \cdot \vec \nabla \left( \frac{\|\vec v\|^2}{2} + w + \phi_b\right) = 0$ but this is obvious from Lamb's equation.
My questions are:
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Why It sufficies to show that $\vec v \cdot \vec \nabla \left( \frac{\|\vec v\|}{2} + w + \phi_b\right) = 0$ to prove that theorem? (The obviousness of the next passage is clear to me)
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How this theorem can be linked to the "classical" Bernoulli statement: $p = -\rho \frac{\|\vec v \|^2}{2} + k$ where $k$ is a constant?
Best Answer
A Streamline (flow line) is a curve with parametric representation $\mathbf{x}_S(s)$ at some fixed time $t$ such that
$$\mathbf{x}_S(s) \times \mathbf{v}(\mathbf{x}_S(s),t) = 0$$
The velocity field is, therefore, tangential to the streamline at any point, and
$$\frac{d \mathbf{x}_S}{ds}(s) = \alpha(s) \mathbf{v}(\mathbf{x}_S(s),t) $$
Defining the quantity $H (\mathbf{x},t) = \frac{\| \mathbf{v}\|^2}{2} + w + \phi_b$, we have
$$\mathbf{v}(\mathbf{x}_S(s),t) \cdot \nabla H(\mathbf{x}_S(s),t) = 0,$$
and, by the chain rule,
$$\frac{\partial}{\partial s} H(\mathbf{x}_S(s),t) = \nabla H(\mathbf{x}_S(s),t)\cdot \frac{d \mathbf{x}_S}{ds}(s) = \alpha(s) \mathbf{v}(\mathbf{x}_S(s),t)\cdot \nabla H(\mathbf{x}_S(s),t) = 0 $$
Therefore, $H$ is constant along a streamline, where value of this constant can be different for each streamline.
What you refer to as the "classical" Bernoulli statement,
$$\tag{1} p + \frac{1}{2} \rho \|\mathbf{v}\|^2 = k,$$
where $k$ is constant at all points in the domain is valid for steady inviscid, incompressible and irrotational flow.
For a proof, begin with the Euler equations governing steady, inviscid flow,
$$\tag{2}\rho (\mathbf{v} \cdot \nabla) \mathbf{v} = - \nabla p,$$
along with the incompressibility condition $\nabla \cdot \mathbf{v} = 0$.
By the general vector identity
$$\nabla (\mathbf{a} \cdot \mathbf{b}) = (\mathbf{a} \cdot \nabla) \mathbf{b} + (\mathbf{b} \cdot \nabla) \mathbf{a} + \mathbf{a} \times (\nabla \times \mathbf{b}) + \mathbf{b} \times (\nabla \times \mathbf{a}),$$
we get with $\mathbf{a} = \mathbf{b} = \mathbf{v}$,
$$\nabla (\|\mathbf{v}\|^2) = \nabla (\mathbf{v} \cdot \mathbf{v}) = 2(\mathbf{v} \cdot \nabla) \mathbf{v} + 2\mathbf{v} \times (\nabla \times \mathbf{v})$$
For irrotational flow, where $\nabla \times \mathbf{v} = 0$, this reduces to
$$\tag{3}(\mathbf{v} \cdot \nabla) \mathbf{v} = \nabla (\frac{1}{2}\|\mathbf{v}\|^2)$$
Since $\rho$ is constant we obtain after substituting (3) into (2),
$$\nabla (\frac{1}{2}\rho\|\mathbf{v}\|^2) = -\nabla p, $$
whence $\nabla (p + \frac{1}{2}\rho\|\mathbf{v}\|^2) = 0$ and (1) must hold where $k$ is a global constant.