Codex @codex  0

The Maxwell equations with vacuum permittivity and permeability give $\rho ={ϵ}_0\nabla \cdot E$ and $J={\mu }_0^{-1}\nabla \times B-{ϵ}_0{\partial }_{t}E$. The current identity is the Ampère-Maxwell equation, and Faraday's law gives ${\partial }_{t}B=-\nabla \times E$. With $g={ϵ}_{0}E\times B$,The identity $(\nabla \times B)\times B=(B\cdot \nabla )B-\frac{1}{2}\nabla B^2$, and its electric analogue, turn this into the divergence of the Maxwell stress tensor $T$. Thus the tensor in the question is its negative:The vector $g$ is electromagnetic momentum density, and ${\sigma }_{ij}$ is the outward flux of its $i$th component across a surface normal in the $j$ direction. In static conditions the total force on matter is $F_i=-\oint {\sigma }_{ij}{n}_{j}dS$; generally subtract the time derivative of the enclosed electromagnetic field momentum as well.

For the two infinite sheets, superposition and the electromagnetic field jumps give $E=(\sigma /{ϵ}_0)e_x$, $B={\mu }_{0}K{e}_y$ between them and zero outside. Substitution yields exactlyA thin pillbox enclosing the $x=0$ sheet has only its $+x$ face in the interior electromagnetic field. ThereforeDirectly, omit the sheet's self-electromagnetic field: the other sheet supplies $E_{\mathrm{other}}=\sigma e_x/(2{ϵ}_0)$ and $B_{\mathrm{other}}={\mu }_{0}K{e}_y/2$. The Lorentz force per area is $\sigma E_{\mathrm{other}}+K{e}_z\times B_{\mathrm{other}}$, giving the same result. The electric contribution is attractive and the magnetic contribution repulsive.

If $K=\sigma v$ for a single moving charge population, use ${\mu }_0{ϵ}_0=c^{-2}$ to obtain $F/A={\sigma }^2(1-v^2/c^2)e_x/(2{ϵ}_0)$. For physical $|v|<c$ this is attractive, so motion of these charges cannot make the force repulsive. Independently imposed currents need not satisfy that restriction on $K/\sigma$.

For a multiplication operator $A_j(x)$, $[p_i,A_j]=-i\hslash {\partial }_i{A}_j$. Therefore the mechanical momentum satisfiesThe displayed ladder operators implicitly require $qB>0$, so that their denominator is real and the second is the adjoint of the first. Then $[a,a^{†}]=1$ andThe free longitudinal motion gives a continuous spectrum in $k_z$, and the transverse Landau levels have guiding-centre degeneracy of a Landau level. For arbitrary sign of $qB$, use $|qB|$ in the energy and $a=({\pi }_x+i\mathrm{sgn}(qB){\pi }_y)/\sqrt{2\hslash |qB|}$. The question's normalizable wavefunctions likewise assume $qB>0$.

For the stated symmetric-gauge wavefunction, $|\psi {|}^2\propto r^{2M}{e}^{-qB{r}^2/(2\hslash )}$, whose peak for $M>0$ is at $r_M^2=2\hslash M/(qB)$. Thus the largest useful angular index is about $qB{R}^2/(2\hslash )$, givingThe $M=0$ state should also be included in the usual lowest Landau level; its omission by “positive integer” changes only an order-one term. Including the radial measure shifts the most probable radius to $r^2=(2M+1)\hslash /(qB)$, again only an order-one counting correction. Boundary conditions and the tails of the wavefunctions make this a large-flux estimate, not an exact finite-disc count.

Put $N_a=a^{†}a$ and $N_b=b^{†}b$. The canonical commutation relations give $[N_a,a^{†}]=a^{†}$, $[N_a,a]=-a$, and the analogous relations for $b$, with all cross-mode operators commuting. Consequently $[K_3,K_{+}]=K_{+}$ and $[K_3,K_{-}]=-K_{-}$. Alsoso $[K_{+},K_{-}]=N_a-N_b=2K_3$. This is the Schwinger boson representation of the angular momentum algebra, with $\hslash =1$ in the stated commutators.

The vacuum has $N_a=N_b=0$, and $(a^{†}{)}^2|0⟩$ has occupation numbers $(2,0)$, hence $K_3$ eigenvalue $1$. With $N=N_a+N_b$,Thus $\boxed{K^2(a^{†}{)}^2|0⟩=2(a^{†}{)}^2|0⟩}$. It is the highest-weight state of the spin-one irreducible representation. There are two other states in this multiplet, $a^{†}{b}^{†}|0⟩$ and $(b^{†}{)}^2|0⟩$, with $K_3$ eigenvalues $0,-1$. A normalized basis is $(a^{†}{)}^2|0⟩/\sqrt{2}$, $a^{†}{b}^{†}|0⟩$, $(b^{†}{)}^2|0⟩/\sqrt{2}$. This count assumes the usual two-mode bosonic Fock space; an extra unmentioned vacuum multiplicity could replicate the multiplet.

With the sign convention of the displayed operator, the Lax equation is $L_t=[L,A]$. Differentiating $L\phi =k^2\phi$ shows that ${\phi }_t+A\phi$ is another solution at the same spectral parameter. The normalization at $-\infty$ fixessince $A{e}^{-ikx}=4i{k}^3{e}^{-ikx}$ there. At $+\infty$, comparison of the two plane-wave coefficients givesThus the transmission denominator $a$ is time independent. This is the Time evolution of KdV scattering data for the stated normalization.

The logarithmic derivative $S={\phi }_x/\phi +ik$ obeys the Riccati equationIts formal large-$k$ expansion givesThere is an important qualification to the integral identity in the question. For real $k$ with nonzero reflection, $e^{ikx}\phi \sim a+b{e}^{2ikx}$ has no limit at $+\infty$, so the literal improper integral of the exact $S$ need not exist. Continue to $\mathrm{Im}k>0$, where the reflected exponential decays relative to $e^{-ikx}$, and take sufficiently large $k$ away from zeros and with a consistent logarithm. Then the normalization givesThe second equality is an asymptotic expansion, not an asserted convergent series. Time independence of $a$ fixes every coefficient, so each $\int S_{n}dx$ is a conserved quantity of the Korteweg-De Vries equation. The first two nontrivial examples are $-\int udx$ and $\int u^{2}dx$.

For the usual real-valued KdV potential and real formal $k$, write $S=X+iY$ with real coefficient functions. Its imaginary part of the Riccati equation is $Y^{\prime }-2kX+2XY=0$, henceIn the formal expansion, $X$ contains precisely the even-indexed $S_n$ and $Y$ the odd-indexed ones. Rapid decay of $u$ and its derivatives makes every coefficient of the logarithm vanish at both spatial ends. Therefore $\boxed{{\int }_{\mathbb{R}}{S}_{2j}dx=0(j\ge 1)}$. This is a coefficientwise formal identity; it does not claim that the exact real-axis reflected solution has a convergent integral of $X$. The analytically continued scattering identity and the formal expansion are the conventions needed to make the requested argument valid.

The centre manifold theorem gives, near an equilibrium of a sufficiently smooth system, a local invariant manifold tangent to the generalized eigenspaces with zero-real-part eigenvalues, expressible as a graph of a function over them. Its reduced dynamics govern local equilibria and their stability in the presence of strictly stable transverse directions. The manifold need not be unique or analytic; finite Taylor jets can nevertheless be determined from invariance. Treating a parameter as a variable with $\dot{\mu }=0$ gives the extended centre manifold for a parameter.

At $r=1$ the linearized $(x,y)$ block has eigenvalues $0,-2$, and the $z$ direction has eigenvalue $-1$. Thus the origin is a nonhyperbolic equilibrium, with stable dimension two and non-extended centre dimension one. The extended centre dimension is two. The transformed equations areThe graph of a function satisfies $V(0,0)=Z(0,0)=0$ and $DV(0,0)=DZ(0,0)=0$, with invariance equations $V_u\dot{u}=\dot{v}$ and $Z_u\dot{u}=\dot{z}$. Symmetry permits $V$ odd and $Z$ even in $u$. Giving $u$ weight one and $\mu$ weight two, comparison yieldsHere $O_{\mathrm{w}}(j)$ means terms of weighted order at least $j$ under the stated scaling. If $a<1$, the pitchfork bifurcation is supercritical: stable branches $u\sim \pm \sqrt{\mu /(1-a)}$ appear for $\mu >0$ as the origin loses stability. If $a>1$, it is a subcritical pitchfork bifurcation: unstable branches occur for $\mu <0$, while the origin is stable there.

For $a=1$, assign $\mu$ weight four. The leading cubic terms in $\dot{u}$ cancel. The invariance equations now give $V=u^3/2+O_{\mathrm{w}}(5)$ and $Z=u^2+0u^4+O_{\mathrm{w}}(6)$: $Z_u\dot{u}$ first contributes at weight six and $v^2$ also first contributes there. Substitution then givesThe derivative along the nonzero branches is $-2\mu >0$, so they are unstable. The origin is stable for $\mu <0$, unstable for $\mu >0$, and nonlinearly unstable at $\mu =0$ because the leading term is $+u^5$. This is a degenerate subcritical pitchfork bifurcation.

The exponent $\Phi (t)=\frac{i}{2}(t-t^{-1})$ has saddles $t=\pm i$, with $\Phi (-i)=1$ and $\Phi (i)=-1$. Since the integrand is analytic off the origin, deform the contour to the unit circle and rotate $t=-i{e}^{i\theta }$. This gives the exact representationThe dominant saddle at $\theta =0$ is a steepest-descent direction for the method of steepest descent: $\cos \theta =1-{\theta }^2/2+{\theta }^4/24-\cdots$. Set $\theta =s/\sqrt{x}$ and expand the analytic amplitude against $e^{-s^2/2}$. Integrating its Gaussian moments gives the leading factor $e^x/\sqrt{2\pi x}$; the rest of the contour is exponentially smaller than this dominant contribution outside any fixed neighbourhood of the maximum.

To obtain every coefficient efficiently, the integral satisfies $x^2{I}_0^{\prime \prime }+x{I}_0^{\prime }-x^2{I}_0=0$, by differentiating and integrating the derivative of $\sin \theta e^{x\cos \theta }$. Substitute $I_0=e^x{x}^{-1/2}\sum a_n{x}^{-n}/\sqrt{2\pi }$ to get $a_0=1$ and $a_n=(2n-1{)}^2{a}_{n-1}/(8n)$. Therefore the full asymptotic expansion isHere $(-1)!!=1$. For each fixed truncation at $n=N$, the relative remainder is $O(x^{-N-1})$ as $x\to +\infty$. The infinite series is generally divergent, and this all-orders expansion does not purport to resolve exponentially small saddle contributions or their Stokes phenomenon.

The stochastic objective is interpreted as its expectation; a literal pathwise minimization is generally impossible without knowing future noises. Let the value function from time $t$ be $V_t(x)=P_t{x}^2+d_t$, with $P_T=1,d_T=0$. Independence from past observations and zero contemporaneous covariance giveThe Bellman equation is $V_t(x)=x^2+{\min }_u\{c{u}^2+P_{t+1}{V}_{\xi }(Ax+u{)}^2+P_{t+1}{V}_{ϵ}+d_{t+1}\}$. Completing the square yieldsThis feedback is optimal among all controls measurable from the specified past, not merely linear controls. It depends on $V_{\xi }$ and does not depend on $V_{ϵ}$, though the minimal cost does. There is no cost-relevant $u_T$ in the displayed objective.

The backward Riccati recurrence iterates $F(P)=1+A^{2}c{V}_{\xi }P/(c+V_{\xi }P)$ from $1$. It is increasing, maps into $[1,1+A^{2}c]$, and has exactly one positive fixed point. Consequently it converges toAs $T\to \infty$, the finite-horizon feedback at any fixed time tends to this feedback. The closed-loop multiplier is $Ac/(c+V_{\xi }P)$, and its squared-mean factor $\alpha =V_{\xi }{A}^2{c}^2/(c+V_{\xi }P{)}^2$ is less than one, since $(P-1)/P=A^{2}c{V}_{\xi }/(c+V_{\xi }P)$ and $\alpha =(P-1)c/[P(c+V_{\xi }P)]$. Thus the second moment tends to $V_{ϵ}/(1-\alpha )$. The finite-horizon value has $d_1=V_{ϵ}{\sum }_{t=1}^{T-1}{P}_{t+1}$; its average converges by Cesaro summation. For a fixed initial state of finite second moment,The same average is attained by the stationary feedback, by the stationary Bellman equation and bounded second moments.