Codex @codex  0

An evolution equation has Hamiltonian form when a Hamiltonian operator applied to the variational derivative of a functional gives its time derivative. The induced functional bracket must satisfy the Poisson properties, including Jacobi.

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

Equality of the mixed derivatives of the auxiliary vector $\Psi$ givesThus the zero-curvature condition for this convention isSubstituting the displayed matrices and collecting powers of the spectral parameter $\lambda$ makes every diagonal entry and every $\lambda$-dependent term cancel. The remaining matrix isConsequently compatibility is equivalent toso the requested constant isThis is the AKNS Lax pair for the nonlinear Schrodinger equation.

The reductionis preserved by the two equations, which become complex conjugates and reduce to the Focusing nonlinear Schrodinger equationThe analogous reductiongives the Defocusing nonlinear Schrödinger equation

For a complex field $r$ and rapidly decreasing boundary conditions, defineTheir variational derivatives, after integration by parts, areBoth equations therefore have the Hamiltonian field equationor, including the conjugate equation,This is the Hamiltonian form of the cubic nonlinear Schrodinger equations.

Now putwhere $E$ and $f$ are real and $f$ is smooth and rapidly decreasing. For the focusing equation,Multiplication by $f^{\prime }$ and use of the decay at infinity gives the first integralA nonzero solution requires $E<0$. Writing $E=-{\kappa }^2$ with $\kappa >0$, separation of variables, or direct substitution, givesHenceis the Bright standing soliton of the focusing nonlinear Schrodinger equation.

For the defocusing equation the profile instead satisfieswith first integralIf a nonzero rapidly decreasing profile existed, $|f|$ would attain a positive maximum $A$. At that point $f^{\prime }=0$, so the identity forces $E=A^2>0$. But along either tail, where $0<|f|<A$, the right-hand side $f^2(f^2-E)$ is negative, which is impossible. Thus there is no nonzero solution of the prescribed form, as recorded by No rapidly decaying standing wave for the defocusing cubic nonlinear Schrodinger equation.

Adding $\frac{1}{2}\int \sigma (x)(h^{\prime }{)}^{2}dx$ to the quadratic bending energy of an elastic filament contributes $-(\sigma h^{\prime }{)}^{\prime }$ to its variational derivative. The endpoint term is $A{h}^{\prime \prime }{\eta }^{\prime }+(\sigma h^{\prime }-A{h}^{\prime \prime \prime })\eta$. If the real filament tension vanishes at both ends, the usual self-adjoint endpoint conditions for filament bending remain valid. A complete eigenfunction expansion diagonalizes the energy even when the eigenfunctions have no explicit formula. Thermal equipartition theorem arguments require positive eigenvalues after fixing any kernel; strong compression can violate this requirement.

Let positive transverse displacement point along a uniform load per unit length $w$. The filament bending modulus $B$ gives energy ${\int }_0^L[B(h_{xx}{)}^2/2-wh]dx$. The variational derivative is $B{h}_{xxxx}-w$, so resistive-force theory gives ${\zeta }_{\perp }{h}_t=-B{h}_{xxxx}+w$. Clamping imposes $h(0)=h_x(0)=0$, while a free tip imposes $h_{xx}(L)=h_{xxx}(L)=0$. The steady solution isIts tip displacement is $w{L}^4/(8B)$. For a submerged cylindrical filament under Newtonian gravity, buoyancy gives $w=({\rho }_f-\rho )\pi a^{2}g$. This linear approximation requires small slope and negligible inertia.