Starting with the Euler equations for an inviscid incompressible fluid, derive Bernoulli's theorem for unsteady irrotational flow.

Inviscid fluid of density $\rho$ is contained within a U-shaped tube with the arms vertical, of height $h$ and with the same (unit) cross-section. The ends of the tube are closed. In the equilibrium state the pressures in the two arms are $p_{1}$ and $p_{2}$ and the heights of the fluid columns are $\ell_{1}, \ell_{2}$.

The fluid in arm 1 is displaced upwards by a distance $\xi$ (and in the other arm downward by the same amount). In the subsequent evolution the pressure above each column may be taken as inversely proportional to the length of tube above the fluid surface. Using Bernoulli's theorem, show that $\xi(t)$ obeys the equation

$$\rho\left(\ell_{1}+\ell_{2}\right) \ddot{\xi}+\frac{p_{1} \xi}{h-\ell_{1}-\xi}+\frac{p_{2} \xi}{h-\ell_{2}+\xi}+2 \rho g \xi=0$$

Now consider the special case $\ell_{1}=\ell_{2}=\ell_{0}, p_{1}=p_{2}=p_{0}$. Construct a first integral of this equation and hence give an expression for the total kinetic energy $\rho \ell_{0} \dot{\xi}^{2}$ of the flow in terms of $\xi$ and the maximum displacement $\xi_{\max }$.