Component proof.

Let $\phi(\vec x)$ be smooth and write components with indices $i,j\in\{1,2,3\}$. The $i$-th component of the left-hand side is

$$\big((\nabla\phi\cdot\nabla)\nabla\phi\big)_i = (\partial_j \phi)\, \partial_j(\partial_i \phi) = (\partial_j \phi)\, \partial_i(\partial_j \phi)$$

(since mixed partials commute for smooth $\phi$). The $i$-th component of the right-hand side is

$$\partial_i\!\left(\tfrac{1}{2}\vert\nabla\phi\vert^2\right) = \pa... ..._k\phi \,\partial_k\phi\right) = (\partial_k\phi)\,\partial_i(\partial_k\phi),$$

which is the same expression (rename index $k$ to $j$). Hence the identity holds:

$$(\nabla\phi\cdot\nabla)\nabla\phi = \nabla\!\left(\tfrac12\vert\nabla\phi\vert^2\right).$$

(b) Derive unsteady Bernoulli for irrotational flow.

We are given an inviscid, incompressible fluid of constant density $\rho$ under gravity with Euler equation

$$\partial_t{\vec u} + (\vec u\cdot\nabla)\vec u = -\frac{1}{\rho}\nabla p - g\hat{\vec z},$$

and assume the flow is irrotational: $\nabla\times\vec u = \vec 0$. Thus there exists a velocity potential $\phi(\vec x,t)$ such that $\vec u=\nabla\phi$.

Substitute $\vec u=\nabla\phi$ into Euler. The left-hand side becomes

$$\partial_t\nabla\phi + (\nabla\phi\cdot\nabla)\nabla\phi.$$

Using part (a) we write

$$(\nabla\phi\cdot\nabla)\nabla\phi = \nabla\!\left(\tfrac12\vert\nabla\phi\vert^2\right).$$

So Euler becomes

$$\partial_t\nabla\phi + \nabla\!\left(\tfrac12\vert\nabla\phi\vert^2\right) = -\frac{1}{\rho}\nabla p - g\hat{\vec z}.$$

Both $\partial_t{t}\nabla\phi$ and the other terms are gradients, so combine into a single gradient:

$$\nabla\!\left(\partial_t{\phi} + \tfrac12\vert\nabla\phi\vert^2 + \frac{p}{\rho} + gz\right) = \vec 0.$$

Thus the scalar inside the gradient is spatially constant (but may depend on time). Hence

$$\partial_t{\phi} + \tfrac12\vert\nabla\phi\vert^2 + \frac{p}{\rho} + gz = f(t),$$

where $f(t)$ is an arbitrary function of time only. This is the unsteady Bernoulli equation for irrotational flow.