[proofplan]
We compute the time derivative of the functional $I[u]$ along a solution of $\partial_t u = \mathcal{J}\, \delta H$ by applying the definition of the functional derivative. The chain rule for functionals expresses $dI/dt$ as the $L^2$ inner product of $\delta I$ with $u_t$. Substituting $u_t = \mathcal{J}\, \delta H$ and recognising the definition of the Poisson bracket $\{I, H\} = \langle \delta I, \mathcal{J}\, \delta H \rangle$ completes the proof in a single computation.
[/proofplan]
[step:Express $dI/dt$ as the $L^2$ pairing of $\delta I$ with $u_t$ via the chain rule for functionals]
Let $u = u(x, t)$ be a solution of $\partial_t u = \mathcal{J}\, \delta H$. By the definition of the time derivative of $I$ along this solution,
\begin{align*}
\frac{dI}{dt}[u] = \lim_{\varepsilon \to 0} \frac{I[u + \varepsilon u_t] - I[u]}{\varepsilon}.
\end{align*}
This is exactly the directional derivative of the functional $I$ in the direction $u_t$. By the definition of the functional derivative $\delta I$, the directional derivative of $I$ at $u$ in the direction $h$ is the $L^2$ pairing $\langle \delta I, h \rangle := \int \delta I \cdot h \, dx$. Applying this with $h = u_t$ gives
\begin{align*}
\frac{dI}{dt} = \langle \delta I, u_t \rangle = \int \frac{\delta I}{\delta u}(x) \, u_t(x) \, dx.
\end{align*}
[/step]
[step:Substitute the equation of motion $u_t = \mathcal{J}\, \delta H$ and identify the Poisson bracket]
Since $u$ satisfies the Hamiltonian PDE, we have $u_t = \mathcal{J}\, \delta H$. Substituting into the expression from the previous step:
\begin{align*}
\frac{dI}{dt} = \langle \delta I, u_t \rangle = \langle \delta I, \mathcal{J}\, \delta H \rangle = \int \frac{\delta I}{\delta u}(x) \left(\mathcal{J}\, \frac{\delta H}{\delta u}\right)(x) \, dx.
\end{align*}
By definition, the Poisson bracket of two functionals $I$ and $H$ with respect to the Hamiltonian operator $\mathcal{J}$ is
\begin{align*}
\{I, H\} := \langle \delta I, \mathcal{J}\, \delta H \rangle = \int \frac{\delta I}{\delta u}(x) \left(\mathcal{J}\, \frac{\delta H}{\delta u}\right)(x) \, dx.
\end{align*}
Therefore
\begin{align*}
\frac{dI}{dt} = \{I, H\},
\end{align*}
which is the claimed identity.
[guided]
This result is the infinite-dimensional analogue of the [Evolution Via Poisson Bracket](/theorems/1334) theorem from finite-dimensional Hamiltonian mechanics: for an observable $f$ evolving under Hamilton's equations $\dot{x} = J \nabla H$, we have $\dot{f} = \{f, H\}$. The structure is identical -- the finite-dimensional gradient $\nabla H$ is replaced by the functional derivative $\delta H$, the symplectic matrix $J$ is replaced by the Hamiltonian operator $\mathcal{J}$, and the finite-dimensional inner product $\langle \nabla f, J \nabla H \rangle$ is replaced by the $L^2$ pairing $\langle \delta I, \mathcal{J}\, \delta H \rangle$.
An immediate consequence is the conservation criterion: a functional $I$ is conserved along the flow if and only if $\{I, H\} = 0$, exactly paralleling the finite-dimensional case. This is the starting point for the systematic search for conserved quantities in integrable PDEs.
[/guided]
[/step]
