= Solution
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 $H=\int h(u,u_x,\ldots)dx$, 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 \b[$\omega=-k^3$], so phase velocity $-k^2$ and group velocity $-3k^2$ depend on wavenumber. This proves small-amplitude <wave dispersion>.
Take
$$
\boxed{H[u]=\frac12\int(u_x^2+u^4)dx,\qquad J=\partial_x}.
$$
Then $\delta H/\delta u=-u_{xx}+2u^3$ and $J\delta H/\delta u=-u_{xxx}+6u^2u_x$, the required flow. On sufficiently smooth functionals define the <Poisson bracket>
$$
\boxed{\{F,G\}=\int\frac{\delta F}{\delta u}\,\partial_x\frac{\delta G}{\delta u}\,dx}.
$$
Variational 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 $dI/dt=\int(\delta I/\delta u)u_tdx=\{I,H\}$. 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 $2u$ gives
$$
\partial_t(u^2)= -2uu_{xxx}+12u^3u_x
=-\partial_x(2uu_{xx}-u_x^2-3u^4).
$$
Integrating the resulting <local conservation law> over the line makes the boundary flux vanish. Hence $I=\int u^2dx$ is conserved and \b[$\{I,H\}=0$]. As a direct check, its <variational derivative> is $2u$, and integration by parts makes $\int2u(-u_{xxx}+6u^2u_x)dx=0$.
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