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Start from $\rho ({\partial }_{t}u+u\cdot \nabla u)=-\nabla p-\rho g{e}_z$, ${\partial }_t\rho +u\cdot \nabla \rho =0$, and $\nabla \cdot u=0$. The stationary background obeys $p_0^{\prime }=-g{\rho }_0$. Linearize in small perturbations and use the Boussinesq approximation: replace the density by a reference value ${\rho }_{\ast }$ in inertia but retain the density perturbation in the buoyancy force. With $\pi =p^{\prime }/{\rho }_{\ast }$ and $b=-g{\rho }^{\prime }/{\rho }_{\ast }$,This is the Brunt–Väisälä frequency. Its assumed constancy requires a locally uniform stable density gradient in this approximation; rotation and viscosity are neglected.

For a plane wave with wavevector $K=(k,\ell ,m)$, project the momentum equation onto the space perpendicular to $K$ to eliminate pressure. Its vertical component is $-i\omega w=b[1-m^2/|K{|}^2]$, while $-i\omega b+N^{2}w=0$. Eliminating $b$ yieldsFor $0<\omega <N$, a small oscillating body radiates upward and downward conical internal gravity wave beams. In a vertical section the beams make a St Andrew's cross, with angle $\alpha$ to the horizontal satisfying $\sin \alpha =\omega /N$. Indeed the group velocity is perpendicular to the wavevector and has ratio of vertical to horizontal magnitudes $\sqrt{k^2+{\ell }^2}/|m|$. Thus wave energy follows the cones, rather than the direction normal to a fixed-radius spherical wavefront. There are no such propagating beams for $\omega >N$.

For the shear-flow calculation choose the positive intrinsic-frequency branch and $k_0>0$, conditions needed for the printed critical-level formula. The slowly varying local dispersion relation isStationarity and horizontal homogeneity conserve $\Omega$ and $k=k_0$. The Hamiltonian ray equations for a local dispersion relation giveThus $m_0<0$ stays negative and the ray propagates upward. Conservation of frequency givesIf the shear reaches the required velocity difference, its first zero is the critical level of an internal gravity wave,As $z\uparrow z_c$, the intrinsic frequency tends to zero, $m\to -\infty$, the vertical wavelength tends to zero and the vertical group velocity tends to zero. The ideal ray cannot cross this level. For finite $U_c^{\prime }>0$, $\dot{z}\sim k_0(U_c^{\prime }{)}^2(z_c-z{)}^2/N$, so it approaches the level only asymptotically in time. The disturbance develops short scales; even weak viscosity can then absorb it, and nonlinear effects can invalidate the small-amplitude approximation.

The printed sign and existence hypotheses are incomplete. On the positive-frequency branch, $k_0<0$ instead gives $\dot{m}>0$ and eventually a turning point when the intrinsic frequency reaches $N$. As a concrete counterexample to the stated universal height bound, take $N=1$, $k_0=-1$, $m_0=-3$, and $U(z)=\epsilon z$ with any small $\epsilon >0$. At the printed height $z_c=1/(\epsilon \sqrt{10})$, the intrinsic frequency is $2/\sqrt{10}<1$ and $m=-\sqrt{3/2}<0$, so the vertical group velocity is still positive and the ray crosses that height. Also, $U^{\prime }>0$ alone does not ensure that the displayed velocity difference is ever reached; an increasing bounded $U$ may have no finite critical level. The qualified derivation above gives the intended positive-$k_0$ result whenever that level exists.

For an axisymmetric flow without azimuthal velocity, define the Stokes streamfunction byThese formulas enforce incompressibility. The only nonzero curl component is $R^{-1}[{\partial }_R(R{u}_{\theta })-{\partial }_{\theta }{u}_R]$. Substitution givesCurling the Stokes flow equations eliminates pressure and gives ${\nabla }^2(\nabla \times u)=0$. For an azimuthal field, its vector Laplacian is the scalar Laplacian minus $1/(R^2{\sin }^2\theta )$; applying that operator to $-D^2\Psi /(R\sin \theta )$ gives $-D^2(D^2\Psi )/(R\sin \theta )$. Hence the streamfunction equation is $\boxed{D^2{D}^2\Psi =0}$.

For the proposed sphere solution put $f(R)=R^2-3aR/2+a^3/(2R)$, so $\Psi =(U/2)f(R){\sin }^2\theta$. Its velocity isBoth components vanish at $R=a$ and tend to the components of $U{e}_z$ at infinity. Also $D^2\Psi =(U/2)(f^{\prime \prime }-2f/R^2){\sin }^2\theta =(3Ua/(2R)){\sin }^2\theta$, and applying $D^2$ once more gives zero. This verifies the differential equation and both physical boundary conditions.

For a sphere translating upward through otherwise stationary fluid, subtract the uniform background in the sphere frame and reverse its far-flow direction. Its leading disturbance is the upward StokesletPlace an image at $z=+d$ with the opposite, downward velocity $-U$, while the real sphere is at $z=-d$. The sum has $u_z$ odd under $z\mapsto -z$ and $u_r$ even. Thus at $z=0$, $u_z=0$ and ${\partial }_z{u}_r=0$; moreover ${\partial }_r{u}_z=0$. The tangential stress ${\sigma }_{rz}=\mu ({\partial }_z{u}_r+{\partial }_r{u}_z)$ is therefore zero, as required by the stress-free boundary condition.

At a point on the plane with radial coordinate $r$, the real Stokeslet gives $u_r=3aUrd/[4(r^2+d^2{)}^{3/2}]$. The image gives the same radial contribution, soThe surface flow is outward. Although the isolated-sphere dipole term is only $O(U(a/d{)}^3)$, the next correction is $O(U(a/d{)}^2)$: the image's velocity at the real sphere is downward and of size $3aU/(4d)$, so satisfying the real sphere's prescribed velocity requires a relative $O(a/d)$ correction to its Stokeslet strength. Since the leading surface flow is $O(Ua/d)$, the first reflection already changes it at the displayed second order. Ignoring this boundary-interaction correction would give the wrong next power.

Use $c=1$ and metric signature $(-,+,+,+)$, consistent with the supplied rest potential. In the Lorenz gauge, the Maxwell equations reduce to wave equations for the four-potential. A point worldline has current $j^a(x)=q\int u^a(s){\delta }^{(4)}(x-y(s))ds$, where $u^a=d{y}^a/ds$. Convolution with the retarded wave Green function $\theta (R^0)\delta (R^2)/(2\pi )$ gives the stated integral: its support is the future light cone of each source point and it excludes incoming advanced signals.

For uniform motion $y^a(s)=(\gamma s,0,0,\gamma vs)$, $u^a=\gamma (1,0,0,v)$. Let $R=x-y(s)$. The delta-function argument has derivative $d(R^2)/ds=-2R\cdot u$. Only the retarded root survives the step function, soPut $Z=z-vt$, $r_{\perp }^2=x^2+y^2$, and $\Delta =t-t_{\mathrm{ret}}>0$. The light-cone condition is ${\Delta }^2=r_{\perp }^2+(Z+v\Delta {)}^2$, whenceTherefore, away from the worldline,As an independent check, in the rest frame $z^{\prime }=\gamma (z-vt)$ and $r^{\prime }=\sqrt{x^2+y^2+{\gamma }^2{Z}^2}$. Its potential is $A^{\prime a}=({\mu }_{0}q/(4\pi r^{\prime }),0,0,0)$. A Lorentz transformation gives $A^0=\gamma A^{\prime 0}$, $A^z=\gamma v{A}^{\prime 0}$, exactly the expression above because $r^{\prime }=\gamma \sqrt{Z^2+(1-v^2)r_{\perp }^2}$. Restoring SI units would require restoring the factors of $c$ in both the four-potential convention and the supplied electrostatic normalization.

Choose the vector potential $A=(-{\int }_0^{y}B(s)ds,0,0)$, whose curl is $(0,0,B(y))$, and a spin-down wavefunction $\Psi =e^{-ikx}\chi (y)(0,1{)}^{\mathsf{T}}$, independent of $z$. DefineThe covariant $x$ derivative acts as $-i(k+a)$, and ${\sigma }_3$ acts as $-1$. The Pauli equation consequently reduces toThe sign of the plane-wave phase was chosen to match the printed $k+a$ convention; choosing $e^{ikx}$ simply reverses the label $k$.

Set $W=k+a$ and $M={\partial }_y+W$. On the square-integrable operator domain with vanishing boundary terms, $M^{†}=-{\partial }_y+W$, andTaking the inner product with $\chi$ gives $ϵ\Vert \chi {\Vert }_2^2=\Vert M\chi {\Vert }_2^2\ge 0$. Thus $\boxed{E\ge 0}$.

A zero mode solves $M\chi =0$, soSince $a(0)=0$ and $a^{\prime }\ge b_0=e{B}_0/\hslash$, integration on either side of zero gives ${\int }_0^{y}a(s)ds\ge b_0{y}^2/2$. Therefore $|{\chi }_k(y)|\le |C_k|e^{-ky-b_0{y}^2/2}$, proving square integrability for every real $k$. Each can be normalized and supplies a distinct momentum label at zero energy. Hence zero energy is infinitely degenerate. In an infinite $x$ direction the individual plane waves are generalized momentum states; square-integrable superpositions of their normalized transverse zero modes give ordinary transverse states by the Fourier isometry. The stipulated zero $z$ momentum is understood per unit longitudinal length, or with periodic longitudinal normalization.

For $t=u+iv$, the exponent isIts imaginary part must remain equal to its value $x$ at the saddle $t=0$. The descending branch connecting the required ends is thereforeAlong it $\sin v=\tanh u$, so the real part is $-x(\cosh u-\mathrm{sech}u)$, strictly negative away from zero for $x>0$. The branch approaches $v=\pm \pi /2$ at the two infinities and has tangent $t=e^{i\pi /4}s$ at the saddle, oriented towards increasing $s$.

Locally $\cosh t=1+t^2/2+O(t^4)$, hence the steepest descent method givesOutside a small saddle neighborhood the descending contour has strictly smaller real exponent; inside it the Gaussian rescaling $s=x^{-1/2}r$ controls the higher terms. Multiplication by $1/(i\pi )$ and taking the real part yieldThis additive asymptotic form specifies the leading term even near the zeros of the oscillating cosine, where a literal pointwise ratio interpretation of the printed $\sim$ is unsuitable.

A standard Brownian motion $W_t$ starts at zero, has continuous sample paths, and has independent increments $W_t-W_s\sim N(0,t-s)$ for $0\le s<t$. The Black-Scholes model assumes a frictionless market permitting continuous, self-financing trading and short selling, a bank account $B_t=e^{rt}$ at constant interest rate, and a non-dividend-paying assetwith constant coefficients and no arbitrage.

For an option price $C(S,t)$, Itô's formula gives diffusion coefficient $\sigma S{C}_S$. Holding $C_S$ units of the asset hedges this diffusion. The hedged portfolio has value $C-S{C}_S$ and instantaneous gain $(C_t+\frac{1}{2}{\sigma }^2{S}^2{C}_{SS})dt$, so no arbitrage requiresEquivalently, the risk-neutral measure makes $e^{-rt}{S}_t$ a martingale and gives the solution $C(S_0,0)=e^{-rT}{\mathbb{E}}_Q(S_T-K{)}^{+}$. Under that measure,Put $d_2=[\log (S_0/K)+(r-{\sigma }^2/2)T]/(\sigma \sqrt{T})$ and $d_1=d_2+\sigma \sqrt{T}$. Then ${\mathbb{P}}_Q(S_T>K)=\Phi (d_2)$, and completing the square in the normal integral gives ${\mathbb{E}}_Q[S_T{1}_{S_T>K}]=S_0{e}^{rT}\Phi (d_1)$. Therefore the Black-Scholes formula isHere $K>0$; a zero strike has value $S_0$ directly.

For the forward-start call option, condition at the strike-fixing time $t$. Given ${\mathcal{F}}_t$, the remaining asset-price ratio has the same lognormal law over $\tau =T-t$, independently of past increments. Thus its time-$t$ value isSince ${\mathbb{E}}_Q(e^{-rt}{S}_t)=S_0$, conditioning once more yieldsThe strike is random before $t$; inserting an unknown $S_t$ as a fixed strike at time zero would miss the conditioning step.

Put $\beta =2/3$. The Bellman equation is $F(x)={\inf }_u\{x^2+u^2+\beta \mathbb{E}F(x+u+ϵ)\}$. For the quadratic candidate $G(x)=P{x}^2+d$, the quantity minimized isFor $P\ge 0$, completing the square gives $u_{\ast }=-\beta Px/(1+\beta P)$. Matching the $x^2$ and constant terms requiresThe first equation is $2P^2-P-3=0$. Its admissible nonnegative root is $P=3/2$, and $d=3$. Hence $\boxed{G(x)=\frac{3}{2}{x}^2+3,u_t=-x_t/2}$.

A suitable discounted verification theorem is: if a nonnegative candidate satisfies the Bellman equation, a stationary minimizing selector has finite cost and its discounted expected terminal candidate tends to zero, and the same terminal limit holds for every finite-cost competing policy, then that selector is optimal and the candidate is the value function. Iterating the Bellman inequality proves the lower bound for each competing cost; iterating equality along the selector proves attainment. Here finite cost implies ${\sum }_t{\beta }^t\mathbb{E}{x}_t^2<\infty$, so ${\beta }^t\mathbb{E}G(x_t)\to 0$. Under the selector $x_{t+1}=x_t/2+{ϵ}_t$, the second moments are bounded by a geometric recursion, giving finite cost and the same terminal limit. Thus $F=G$. Infinite-cost policies cannot improve the value.

For $\lambda =0$, the two coordinates decouple and have the same discount, so $\boxed{u_t=-x_t/2,w_t=-y_t/2}$. The total value is $(3/2)(x^2+y^2)+6$.

For $\lambda =1$, the discount is $\beta =3/4$ and the state cost is $(x+y{)}^2$. Introduce the orthogonal transformation $s=(x+y)/\sqrt{2}$, $q=(x-y)/\sqrt{2}$, with controls $v=(u+w)/\sqrt{2}$, $z=(u-w)/\sqrt{2}$. The cost becomes $2s^2+v^2+z^2$. Each transformed noise has mean zero and variance one; they need not be independent unless the original noises are Gaussian, and no such independence is needed here. The unpenalized difference coordinate has optimal control $z=0$. For the sum coordinate, the quadratic coefficient solvesso $P=8/3$, $v=-2s/3$, and the constant is $d=\beta P/(1-\beta )=8$. ThereforeThe sum state is mean-square stable under this feedback; the uncontrolled difference state may wander, but its value function and state cost are both zero. The same verification argument applies to the penalized sum coordinate.