What does it mean for an evolution equation $u_{t}=K\left(x, u, u_{x}, \ldots\right)$ to be in Hamiltonian form? Define the associated Poisson bracket.

An evolution equation $u_{t}=K\left(x, u, u_{x}, \ldots\right)$ is said to be bi-Hamiltonian if it can be written in Hamiltonian form in two distinct ways, i.e.

$$K=\mathcal{J} \delta H_{0}=\mathcal{E} \delta H_{1}$$

for Hamiltonian operators $\mathcal{J}, \mathcal{E}$ and functionals $H_{0}, H_{1}$. By considering the sequence $\left\{H_{m}\right\}_{m \geqslant 0}$ defined by the recurrence relation

$$\mathcal{E} \delta H_{m+1}=\mathcal{J} \delta H_{m}$$

show that bi-Hamiltonian systems possess infinitely many first integrals in involution. [You may assume that $(*)$ can always be solved for $H_{m+1}$, given $H_{m}$.]

The Harry Dym equation for the function $u=u(x, t)$ is

$$u_{t}=\frac{\partial^{3}}{\partial x^{3}}\left(u^{-1 / 2}\right)$$

This equation can be written in Hamiltonian form $u_{t}=\mathcal{E} \delta H_{1}$ with

$$\mathcal{E}=2 u \frac{\partial}{\partial x}+u_{x}, \quad H_{1}[u]=\frac{1}{8} \int u^{-5 / 2} u_{x}^{2} \mathrm{~d} x$$

Show that the Harry Dym equation possesses infinitely many first integrals in involution. [You need not verify the Jacobi identity if your argument involves a Hamiltonian operator.]