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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 $\int 2u(-u_{xxx}+6u^2{u}_x)dx=0$.

The James–Stein estimator in the specified unit-variance model iswith an arbitrary value at $X=0$, a probability-zero event. The risk of $X$ is $p$. Gaussian Stein identity states $E_{\theta }[(X_i-{\theta }_i)g_i(X)]=E_{\theta }[{\partial }_i{g}_i(X)]$ for suitably integrable differentiable functions; it extends to the present singular function by truncation, since $p>2$ makes $E\Vert X{\Vert }^{-2}$ finite.

Put $c=p-2$ and $g(x)=-cx/\Vert x{\Vert }^2$. Then $\Vert g{\Vert }^2=c^2/\Vert x{\Vert }^2$ and $\mathrm{div}g=-c(p-2)/\Vert x{\Vert }^2$. Expanding the quadratic risk and applying Stein's lemma yieldsThus domination is strict for every finite $\theta$.

To identify the worst-case risk, let $r=\Vert \theta \Vert \to \infty$. On $\Vert X\Vert >r/2$, the reciprocal square is at most $4/r^2$. On the complementary ball, the Gaussian density is at most $(2\pi {)}^{-p/2}{e}^{-r^2/8}$, so its reciprocal-square integral is bounded by a constant times $e^{-r^2/8}{r}^{p-2}$, tending to zero. Therefore $E_{\theta }\Vert X{\Vert }^{-2}\to 0$, and $R(\theta ,\hat{\theta })\to p$ from below. HenceThe second supremum is approached at infinity and need not be attained at a finite parameter.

The planar planar isoperimetric inequality iswith equality precisely for a disc. Let the positively oriented boundary be parametrized by arclength as $r(s)=(x(s),y(s))$, $0\le s\le L$. Translate its mean to zero. Then $|r^{\prime }|=1$, and Green's theorem gives $2A={\int }_0^L(x{y}^{\prime }-y{x}^{\prime })ds$.

The periodic Wirtinger inequality says that for a real absolutely continuous $L$-periodic function $h$ of mean zero and square-integrable derivative,with equality exactly for $h(s)=a\cos (2\pi s/L)+b\sin (2\pi s/L)$. Apply it to both coordinates and use Cauchy-Schwarz inequality:If equality holds, the Wirtinger equalities give $r=p\cos (2\pi s/L)+q\sin (2\pi s/L)$. The arclength condition forces $|p|=|q|=L/(2\pi )$ and $p\cdot q=0$, so the boundary is a circle. A circle attains equality, making the constant optimal.

The inequality and equality characterization persist for finitely many corners. A rectifiable piecewise smooth simple closed curve has a Lipschitz arclength parametrization, with $|r^{\prime }|=1$ almost everywhere. Green's theorem and the absolutely continuous version of Wirtinger's inequality used above still apply. More generally one may approximate a rectifiable Jordan curve by smooth curves while preserving the limiting length and area. If the allowed isolated singularities give infinite length, the inequality is trivially satisfied in the extended sense.