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Substitution of $e^{i(kx-\omega t)}$ gives the dispersion relation $\boxed{\omega (k)=Uk+k^3}$. The phase velocity is $U+k^2$, whereas the group velocity is $U+3k^2$. Thus nonzero-wavenumber crests move more slowly than their wave packet; the velocities coincide at zero wavenumber.

Fourier evolution gives $\varphi (Vt,t)=\int A(k)e^{it\Phi (k)}dk$, where $\Phi (k)=(V-U)k-k^3$. For $V>U$ the two nondegenerate stationary points are $k=\pm \kappa$, $\kappa =\sqrt{(V-U)/3}$. At these points $\Phi (\pm \kappa )=\pm 2{\kappa }^3$ and ${\Phi }^{\prime \prime }(\pm \kappa )=\mp 6\kappa$. Expanding the phase quadratically near each point and evaluating the oscillatory Gaussian integral gives the stationary phase methodFor a real initial field, $A(-\kappa )=\overline{A(\kappa )}$, so the two terms are conjugate. The approximation assumes a sufficiently smooth decaying amplitude; if a leading coefficient vanishes, the next stationary-phase terms determine that contribution.

For $V<U$, ${\Phi }^{\prime }(k)=V-U-3k^2<0$ everywhere. There are no real stationary points, so no leading propagating packet reaches such a ray. For a Schwartz amplitude, repeated integration by parts with $e^{it\Phi }=(it{\Phi }^{\prime }{)}^{-1}{\partial }_k{e}^{it\Phi }$ makes the contribution $O(t^{-N})$ for every fixed $N$. With weaker data the decay rate depends on their regularity. This agrees with the minimum group velocity being $U$.

The incompressible Navier-Stokes equations are $u_t+(u\cdot \nabla )u=-\nabla p/\rho +\nu \Delta u$ and $\nabla \cdot u=0$, with no penetration and no slip at a stationary rigid wall. The Euler limit takes $\nu \to 0$ at fixed distance from the wall, leaving the inviscid Euler equations and a no-penetration condition; imposing tangential no slip on that outer problem generally overdetermines it. The Prandtl limit resolves a shrinking wall layer in which transverse viscous diffusion remains comparable to convection and restores no slip.

For the steady flat-plate boundary layer, $\delta /x\ll 1$, continuity makes $v/U=O(\delta /x)$, and $u_{xx}$ is small compared with $u_{yy}$. Normal momentum makes pressure independent of $y$ to leading order. The constant outer speed makes the outer pressure gradient, and hence $p_x$ throughout the layer, zero. The leading equations are $u{u}_x+v{u}_y=\nu u_{yy}$ and $u_x+v_y=0$. With the stream function convention $u={\psi }_y$, $v=-{\psi }_x$, these giveBalancing $U^2/x$ against $\nu U/{\delta }^2$ gives $\delta =\sqrt{\nu x/U}$. There is no independent streamwise scale for the semi-infinite plate, motivating the similarity solutionIt gives $u=U{f}^{\prime }$ and $v=\frac{1}{2}\sqrt{\nu U/x}(\eta f^{\prime }-f)$. Substitution yields $u{u}_x+v{u}_y=-U^{2}f{f}^{\prime \prime }/(2x)$ and $\nu u_{yy}=U^2{f}^{\prime \prime \prime }/x$, so the Blasius equation and conditions areThe first two conditions impose no penetration and no slip; the last matches the outer stream.

The wall shear on one face is $\tau (x)=\rho \nu u_y(x,0)=\rho U\sqrt{\nu U/x}{f}^{\prime \prime }(0)$. Integration from zero to $L$ gives $2\rho U^{2}L{f}^{\prime \prime }(0)/\sqrt{UL/\nu }$ on one face. With the same flow on both faces, the total drag force per unit width isOur normalization has $f^{\prime }(\infty )=1$, so the displayed denominator is one. The factor four refers to both faces; for a plate wetted on only one face the drag is half this value.

Choose polar angle $\theta$ from the acceleration direction. Then $|n\times q\dot{v}{|}^2=q^2|\dot{v}{|}^2{\sin }^2\theta$. Integrating over the solid angle gives $2\pi {\int }_0^{\pi }{\sin }^3\theta d\theta =8\pi /3$, and hence the specified normalization gives Larmor formulaThis uses the source's convention for radiated power; standard SI normalization has an additional factor $1/c$ in both given power formulas.

To leading order when $W\ll E$, neglect the loss while computing the head-on trajectory. Mechanical energy conservation gives $\frac{1}{2}m{\dot{r}}^2+V(r)=E=V(r_0)$ and acceleration magnitude $|V^{\prime }(r)|/m$. On each leg,The incoming and outgoing legs contribute equally. Multiplying the power by this time element, adding both legs, and assuming the integral converges givesUsing the unperturbed turning point is justified at this order in the small radiative loss, for a regular turning trajectory.

Periodic boundary conditions give wavevectors $(2\pi /L)(n_x,n_y)$. In the continuum, large-box limit each state occupies area $(2\pi /L{)}^2$ in wavevector space. A circular annulus therefore contains $[L^2/(2\pi {)}^2]2\pi kdk$ spinless states. Since $ϵ={\hslash }^2{k}^2/(2m)$, the density of states isAt finite $L$ the exact spectrum is discrete; this formula is its continuum density.

With electron spin the two Zeeman branches are ${ϵ}_{\mathrm{kin}}\mp \mu H$. The requested calculation includes this spin coupling and neglects orbital coupling to the field; for a perpendicular field, Landau levels would change the result. It is appropriate for a field parallel to the ideal two-dimensional plane, or for the stated spin-only model. Taking $\mu H\ge 0$, the lower branch starts at $-\mu H$, and the upper one at $\mu H$, soConstruct the ground state by filling the lowest one-electron levels subject to the Pauli exclusion principle. If the Fermi energy ${ϵ}_F$ exceeds $\mu H$, the two populations are $N_{+}=g_0({ϵ}_F+\mu H)$ and $N_{-}=g_0({ϵ}_F-\mu H)$. ConsequentlyThe ground-state energy, including the Zeeman term, follows by integrating the energy over each occupied branch:

A Bravais lattice is the set of all integer combinations of linearly independent primitive translation vectors. Its reciprocal lattice consists of $g$ satisfying $g\cdot l\in 2\pi \mathbb{Z}$ for every translation $l$; equivalently $e^{ig\cdot l}=1$.

The original PDF's third family has a bare $j$, not $n_{2}j$. Taken literally, the listed set is not the claimed Bravais lattice: the claimed primitive vector $a(i+k)/2$ does not occur. Its half-integer first and third coordinates force it into that third family, whose second coordinate is always $a$, not zero. The intended repair is $j\mapsto n_{2}j$ in the third family. The following conclusions apply to that corrected set.

The four corrected families are exactlyThe allowed parity patterns are all even, or exactly two odd coordinates. An integer combination of the proposed primitive vectors has coordinates $(a/2)(m_2+m_3,m_1+m_3,m_1+m_2)$, so lies in this set. Conversely, solveAll three are integers precisely when $h+k+l$ is even. This proves that the corrected set is the face-centered cubic lattice with the proposed primitive basis.

Direct dot products give $a_i\cdot b_j=2\pi {\delta }_{ij}$ forThese vectors generate the entire reciprocal lattice, since the coefficients of an arbitrary reciprocal vector in this basis are its integer pairings with the $a_i$. Its vectors are $(2\pi /a)(h,k,l)$ with all three coordinates of the same parity, the body-centered cubic lattice.

In Bragg scattering, translating a scattering atom by $l$ changes the amplitude by $e^{i(k-k^{\prime })\cdot l}$. Constructive interference from every lattice translate requires this factor to be one, so $k^{\prime }-k=g$ is a reciprocal lattice vector. For elastic scattering, $|k^{\prime }|=|k|$; the chord between these two vectors has length $2|k|\sin (\theta /2)$. ThereforeThe shortest nonzero reciprocal vectors are the $(111)$ family of length $(2\pi /a)\sqrt{3}$, followed by the $(200)$ family of length $(2\pi /a)2$. Identical atoms on this Bravais lattice introduce no additional basis extinction. If both families are accessible at the incident wavelength, their scattering angles obey $\boxed{\sin ({\theta }_1/2)/\sin ({\theta }_2/2)=\sqrt{3}/2}$.

Addition of two spin-one angular momenta gives $J=0,1,2$, with $M=-J,-J+1,\dots ,J$. To obtain the $J=1$ multiplet, its highest-weight state must have $M=1$ and be killed by $J_{+}=J_{+}^{(1)}+J_{+}^{(2)}$. On the two possible product states, the raising operator gives the same multiple of $|1;1⟩$, so their difference is required. Normalization and successive lowering giveFor example $J_{-}|1,1⟩=|1;-1⟩-|-1;1⟩=\sqrt{2}|1,0⟩$, and lowering once more gives the last expression. These states are all antisymmetric under interchanging the two constituents.

Identical spin-one particles are bosons, so the total state must be symmetric. Its $J=1$ spin factor is antisymmetric and must therefore multiply an antisymmetric spatial factor. Two particles in the same one-particle orbital have zero antisymmetric spatial state. The lowest allowed choice isFor distinguishable particles there is no exchange-symmetry constraint. Both can occupy the ground orbital, giving ${\psi }_1(r_A){\psi }_1(r_B)|1,M⟩$ and lowest energy $\boxed{2E_1}$, again with three spin states.

Watson lemma expands a Laplace integral by integrating the small-$t$ expansion term by term. More precisely, if $f(t)\sim {\sum }_{j\ge 0}{a}_j{t}^{\lambda +j-1}$ with $\mathrm{Re}\lambda >0$, and the tail satisfies bounds permitting this localization, thenFor ordinary powers $f(t)\sim \sum c_j{t}^j$, the coefficients are $c_{j}j!$. For the Gaussian weight, put $s=t^2$ separately on the positive and negative half-lines. The integral becomes a Laplace integral with amplitude $[f(\sqrt{s})+f(-\sqrt{s})]/(2\sqrt{s})$. Odd terms cancel and Watson lemma givesThese assertions concern asymptotic expansions, not necessarily convergent series.

For the complex integral in the lower half-plane, the identity $(z-t{)}^{-1}=i{\int }_0^{\infty }{e}^{-i(z-t)s}ds$ and the Gaussian Fourier transform giveThe last expression, initially obtained by completing the square, is entire and therefore gives the requested analytic continuation. Expanding the denominator formally, with odd Gaussian moments zero, gives the algebraic expansionHere $(-1)!!=1$. The complementary error function expansion shows $I(z)\sim \mathcal{A}(z)$ on closed subsectors of $-5\pi /4<\arg z<\pi /4$, using an unwrapped argument centered on the lower half-plane.

In the upper half-plane, the directly evaluated real-contour integral is $J(z)=-i\pi e^{-z^2}\mathrm{erfc}(-iz)$. Since $\mathrm{erfc}(w)+\mathrm{erfc}(-w)=2$, continuation from below givesThe upper expansion for $J$ holds on closed subsectors of $-\pi /4<\arg z<5\pi /4$. The added term is the residue acquired by the continued contour passing the pole; it is exponentially dominant in the upper sector $\pi /4<\arg z<3\pi /4$ and exponentially small near either real ray. On the real axis the continued value is the principal-value integral plus $i\pi e^{-x^2}$.

To make the Stokes phenomenon convention explicit, the exponential switches its asymptotic multiplier across the real rays $\arg z=0,\pi$; these are the Stokes switching lines. The diagonal rays $\arg z=\pi /4+j\pi /2$ separate exponential growth from decay and are often called anti-Stokes lines, or critical boundaries when regions are classified by dominance. Thus both sets of critical directions and the exact jump are specified, independently of which terminology is used. This is the Gaussian Cauchy transform and its Stokes jump.

For positive time, convolution with the heat kernel givesDifferentiation under the integral verifies the heat equation, and the approximate identity property gives the initial value. Taking absolute values under the integral and using $e^{-|x-y{|}^2/(4t)}\le 1$ givesThe printed inequality omits the modulus even though $g$ is complex-valued. A complex number has no compatible order, so the displayed absolute-value estimate is the meaningful correction.

Substitution of $u=e^{it}v$ reduces the forced partial differential equation to $(i-\Delta )v=f$. With the Fourier transform convention for which $\widehat{\Delta v}=-|\xi {|}^2\hat{v}$, setThe denominator never vanishes for real $\xi$. Every derivative of its reciprocal has at most polynomial growth, so multiplying a Schwartz function by it gives another Schwartz function. The inverse Fourier transform therefore supplies the required $v$.

For the prescribed initial value the full solution is $u(t,x)=e^{it}v(x)+K_t\ast (g-v)(x)$. Applying the heat kernel estimate to the Schwartz function $g-v$ proves the stronger uniform conclusion

Interpret unit-intensity Gaussian white noise as $d{W}_t/dt$. Differentiating $z=x+(T-t)y$ cancels the two $ydt$ terms and givesThe cost is terminal $z(T{)}^2$ plus the same running $u^2$. Write $h=T-t$ and guess the value $V(t,z)=P(t)z^2+C(t)$. The Hamilton-Jacobi-Bellman equation isCompleting the square gives $u_{\ast }=-hPz$. Equating powers yields $P^{\prime }=h^2{P}^2$, $C^{\prime }=-h^{2}P$, with $P(T)=1,C(T)=0$. Integration givesIndeed $V_t+{\mathcal{L}}^{u}V+u^2=(u+hPz{)}^2\ge 0$, and the Itô verification identity attains equality at this control. This proves optimality for admissible controls of finite expected quadratic cost. The additive noise changes only $C$, not the feedback gain, so this is certainty-equivalence control. A different constant noise intensity multiplies $C$ but leaves the control unchanged, as in the terminal-position reduction of double-integrator control.