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A Hamiltonian evolution equation has the form $u_t=J\delta H/\delta u$, where $H$ is a functional and the Hamiltonian operator $J$ induces a bilinear antisymmetric bracket obeying the Jacobi identity. For a local functional $H=\int h(u,u_x,\dots )dx$, its variational derivative is the coefficient of a test perturbation after all integrations by parts.

Linearizing about zero drops the cubic term and gives $u_t+u_{xxx}=0$. A plane wave $e^{i(kx-\omega t)}$ has $\omega =-k^3$, so phase velocity $-k^2$ and group velocity $-3k^2$ depend on wavenumber. This proves small-amplitude wave dispersion.

TakeThen $\delta H/\delta u=-u_{xx}+2u^3$ and $J\delta H/\delta u=-u_{xxx}+6u^2{u}_x$, the required flow. On sufficiently smooth functionals define the Poisson bracketVariational differentiation and integration are linear, proving linearity in both arguments. Integration by parts and the rapid decay give $\{F,G\}=-\{G,F\}$. The constant skew differential operator ${\partial }_x$ also satisfies the Hamiltonian Jacobi condition, so this is indeed a Poisson bracket.

Along the flow, the functional chain rule gives $dI/dt=\int (\delta I/\delta u)u_{t}dx=\{I,H\}$. Thus an autonomous functional is a first integral exactly when its bracket with $H$ vanishes identically on phase space. Direct multiplication of the equation by $2u$ givesIntegrating the resulting local conservation law over the line makes the boundary flux vanish. Hence $I=\int u^{2}dx$ is conserved and $\{I,H\}=0$. As a direct check, its variational derivative is $2u$, and integration by parts makes $\int 2u(-u_{xxx}+6u^2{u}_x)dx=0$.