An unsteady incompressible inviscid fluid flow satisfies the continuity equation $∇·\textbf u = 0$ and Euler’s equation $$\frac{∂ \textbf u }{∂ t} +(\textbf u·∇)\textbf u = − \frac1ρ ∇p$$ where $\textbf u$ is the velocity, $t$ is time, $ρ$ is the constant density and $p$ is pressure.
(i) If the flow is also irrotational, show that the potential satisfies Laplace’s equation where the potential is defined by $u = ∇ φ$.
(ii) Use Euler’s equation and the vector identity $$∇(\textbf a·\textbf b) = (\textbf a·∇)\textbf b+(\textbf b·∇)\textbf a+\textbf a∧(∇∧\textbf b)+\textbf b∧(∇∧\textbf a)$$ to show that $$\frac{∂φ }{∂ t} + \frac12 |∇ φ |^2 + \frac pρ = \text{constant}$$
Is the first part simply $ \nabla \cdot \textbf u =0 = \nabla \cdot \nabla φ= \nabla ^2 φ =0$?
For the second part, I have tried for ages and can't seem to get it, are we allowed to cancel the grad from both sides. And I have no idea where the half comes from.
Using the identity (ii) we have
$$∇(\textbf u·\textbf u) = 2(\textbf u·∇)\textbf u + 2\textbf u∧(∇∧\textbf u).$$
In irrotational flow , $∇∧\textbf u = 0$ and $\textbf u = \nabla \phi$.
Hence,
$$ 2(\textbf u·∇)\textbf u =∇\left(\textbf u·\textbf u\right) = ∇\left(|\nabla \phi|^2\right) \\ \implies (\textbf u·∇)\textbf u = ∇\left(\frac{1}{2}|\nabla \phi|^2\right).$$
Substitution in Euler's equation yields
$$\frac{\partial}{ \partial t} \nabla \phi + ∇\left(\frac{1}{2}|\nabla \phi|^2\right) + ∇\left(\frac{p}{\rho}\right)= 0 \\ \implies \nabla H := \nabla \left(\frac{\partial}{ \partial t} \phi + \frac{1}{2}|\nabla \phi|^2 + \frac{p}{\rho}\right)= 0.$$
Integrating each component of $\nabla H = 0$ we obtain the Bernoulli equation
$$\frac{\partial}{ \partial t} \phi + \frac{1}{2}|\nabla \phi|^2 + \frac{p}{\rho} = C(t),$$
where "constant" $C$ is independent of position but may depend on time.