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For a nonzero wavevector $\mathbf{k}$, substitute the plane wave into part (c). Each time derivative contributes $-i\omega$, each Laplacian contributes $-|\mathbf{k}{|}^2$, and each second $z$ derivative contributes $-k_z^2$. A nonzero amplitude therefore requiresIf $\theta$ is the angle between the wavevector and the rotation axis, $\cos \theta =k_z/|\mathbf{k}|$, andThis is the inertial wave dispersion relation. Incompressibility also requires $\mathbf{k}\cdot {\mathbf{U}}_0=0$; the original first-order equations select the corresponding transverse polarizations. A horizontal wavevector gives the zero-frequency geostrophic case.

Take the curl of the linear momentum equation. For constant $\mathbf{\Omega }$ and zero divergence, $\nabla \times (\mathbf{\Omega }\times \mathbf{u})=-(\mathbf{\Omega }\cdot \nabla )\mathbf{u}$. With $\mathbf{\Omega }=\Omega {\mathbf{e}}_z$, this givesTake another curl and use part (a): $-{\partial }_t{\nabla }^2\mathbf{u}=2\Omega {\partial }_z\mathbf{\omega }$. Differentiate in time and substitute the previous equation:ConsequentlyThe two curl operations remove the pressure gradient while retaining the rotational restoring force.

For constant density $\rho$ and constant angular velocity $\mathbf{\Omega }$, the Euler equations for an inviscid fluid in the rotating frame isThe last two forces are gravity and centrifugal acceleration; $2\mathbf{\Omega }\times \mathbf{u}$ is the Coriolis force term on the left. Incorporate both conservative forces into the modified pressureso the right side is $-\nabla \Pi$.

At latitude $\lambda$ on a spherical Earth of radius $R$, the outward local vertical component of centrifugal acceleration is ${\Omega }_E^{2}R{\cos }^2\lambda$. Its ratio to gravity is at most ${\Omega }_E^{2}R/g\simeq 3.5\times 10^{-3}$, using ${\Omega }_E\simeq 7.3\times 10^{-5}{\mathrm{s}}^{-1}$, $R\simeq 6.4\times 10^6\mathrm{m}$, and $g\simeq 9.8{\mathrm{m}\mathrm{s}}^{-2}$. Thus this component is small compared with gravity everywhere on Earth.

For a velocity scale $U$ and length scale $\ell$, nonlinear advection scales as $U^2/\ell$, whereas the Coriolis term scales as $2\Omega U$. Their ratio is the Rossby number $U/(2\Omega \ell )$. If it is small, for example by increasing rotation at fixed $U,\ell$, neglect nonlinear advection in perturbations about a fluid at rest. The conservative equilibrium forces have already been absorbed into pressure; with a rotation-timescale unsteady motion the leading equations areThese are the linear equations governing inertial waves.

By definition the vorticity is $\mathbf{\omega }=\nabla \times \mathbf{u}$. The curl of the curl identity yieldsFor incompressible flow, $\nabla \cdot \mathbf{u}=0$, so

Write the pole as $(x_0,y_0)$ with $x_0,y_0>0$. The method of images uses three reflected poles with signs $-,-,+$. ThusThe logarithms cancel in pairs on each axis. Only the original pole is in the interior, so this is the required Dirichlet Green function in a quadrant.

On the horizontal boundary $(t,0)$ the outward normal is $-{\mathbf{e}}_y$. Differentiating the four logarithms givesMultiply by the horizontal Dirichlet boundary condition $e^{-t^2}$ and integrate to obtainBy exchanging the axes, the vertical boundary contributes $F(y_0,x_0)$. The representation in part (a) consequently givesFor the unbounded-domain condition, the same expression is the Poisson integral after the conformal change of variables $w=(x+iy{)}^2$ to the upper half-plane, with real-axis data $e^{-|s|}$. This continuous boundary function tends to zero at both ends. Its Poisson integral tends to zero as $|w|\to \infty$: split the data into a compact part, whose kernel integral tends to zero, and a tail uniformly smaller than any prescribed positive number. Hence the constructed function has the required decay, and its two boundary traces agree at the corner. The maximum principle for harmonic functions on expanding quarter-discs proves uniqueness among decaying solutions. The corner value is obtained by continuity as $1$.

With the positive logarithmic fundamental solution used here, a Dirichlet Green function for an interior pole ${\mathbf{r}}_0$ satisfiesEquivalently, $G-G_0$ is harmonic near the pole, $G$ is harmonic away from it, and it has the specified logarithmic singularity. Regularity up to the remaining boundary is understood; on an unbounded domain impose the appropriate condition at infinity as well.

Apply Green second identity with $f=\psi$ and $g=G$. Since $\psi$ is a harmonic function, the volume integral is $\int \psi {\delta }_{{\mathbf{r}}_0}dA=\psi ({\mathbf{r}}_0)$. The term $G{\partial }_n\psi$ vanishes on the boundary by the Dirichlet boundary condition. ThereforeOne may equivalently excise a small disc around the pole and take its radius to zero. The sign is positive because this question uses ${\nabla }^{2}G=+\delta$, not the alternative negative-Laplacian convention.

Here $b>0$ because $a=0$ and the two cost coefficients are not both zero. The interior Euler-Lagrange equation makes $V$ quadratic. Imposing each displayed pair of endpoint data and the distance constraint gives the following formal profiles and their costs:Thus choice (3) is the best strategy, and its true minimum cost isThe qualification from part (b) applies to row (1): its formal quadratic is not a minimizer with free terminal speed, since $\dot{V}(T)\ne 0$. Row (2) is a genuine minimizer for its fixed zero terminal speed, and row (3) is the genuine free-terminal-speed minimizer, satisfying its natural condition $\dot{V}(T)=0$.

A direct lower bound makes the comparison rigorous. For every admissible speed with $V(0)=0$,by integration by parts. The Cauchy-Schwarz inequality gives $L^2\le (T^3/3){\int }_0^T{\dot{V}}^{2}dt$. Equality requires $\dot{V}$ to be proportional to $T-t$, giving exactly row (3). Row (1)'s zero initial acceleration excludes equality; however short initial layers, as in part (b), approach that same infimum. Hence choice (1) has infimum $3b{L}^2/T^3$ but does not attain it, while choice (3) does. This distinction resolves the boundary-condition issue in the printed minimization request.

Put $\kappa =\sqrt{a/b}>0$. Solving the interior Euler-Lagrange equation and imposing both $V(0)=0$ and $\dot{V}(0)=0$ yields $V=C(\cosh \kappa t-1)$, with $\lambda =-2aC$. The distance constraint fixesThe denominator is positive because $\sinh z>z$ for $z>0$. This is the requested solution of the differential equation with the stated initial conditions.

As a minimum-energy problem with free terminal speed, the printed conditions do not admit a smooth minimizer. The formal profile has $\dot{V}(T)=C\kappa \sinh (\kappa T)>0$, violating the natural terminal condition from part (a). Fixing the initial acceleration restricts ${\eta }^{\prime }(0)$, but it does not restrict the independently variable $\eta (T)$ and therefore does not remove that condition.

For clarity, without the extra initial-acceleration restriction the actual unique minimizing profile isIt satisfies $V_{\ast }(0)=0$, ${\dot{V}}_{\ast }(T)=0$, the distance constraint and the interior equation, but ${\dot{V}}_{\ast }(0)>0$. For any perturbation preserving the distance and initial speed, the first variation about $V_{\ast }$ is zero, andwith equality only for $\eta =0$. Its energy is $a{L}^2/[T-\tanh (\kappa T)/\kappa ]$.

This same energy is the unattained infimum under the extra condition $\dot{V}(0)=0$. Indeed multiply $V_{\ast }$ by $1-e^{-t/\epsilon }$ and rescale by a factor tending to one to restore total distance $L$. These smooth nonnegative profiles have both initial values zero and converge to $V_{\ast }$ in value and derivative square integrals as $\epsilon \downarrow 0$. Their energies tend to $E[V_{\ast }]$, but uniqueness of $V_{\ast }$ precludes attaining it with zero initial acceleration. This is nonattainment under an initial derivative constraint.

Take $T>0$ and a positive prescribed distance $L$; if $L=0$, the zero-speed journey has zero cost. Minimize $E[V]={\int }_0^T(a{V}^2+b{\dot{V}}^2)dt$ subject to $V(0)=0$ and the isoperimetric constraint ${\int }_0^{T}Vdt=L$. Introduce a constant Lagrange multiplier $\lambda$ and the integrand $a{V}^2+b{\dot{V}}^2-\lambda V$. For a variation $\eta$ with $\eta (0)=0$, integration by parts givesThe interior Euler-Lagrange equation isWhen terminal speed is free, the same calculation also imposes the natural boundary conditions for a free endpoint:One cannot drop this boundary term when interpreting an extremal as a genuine minimum. If $b=0<a$, the interior condition makes $V$ constant; with a continuous zero initial speed and $L>0$ it cannot be attained. In that case Cauchy-Schwarz inequality gives the infimum $a{L}^2/T$, approached by speeds nearly constant except in a short initial layer. For $b>0$, the derivative cost controls such layers and the natural terminal condition matters in parts (b) and (c).

Constant Gaussian curvature one gives $f^{\prime \prime }+f=0$, so, after a shift of the arc-length parameter, $f=C\cos u$ with $C>0$ on an interval contained in $(-\pi /2,\pi /2)$. To close the surface by adding only two points, the limiting parallels at both ends must collapse to points on the axis. If an end had limiting radius $f>0$, it would contribute a whole missing circle, which two added points cannot supply. Thus the parameter interval is the full interval between consecutive zeros, and its two ends are $u=\pm \pi /2$.

At a smooth axis point the tangent plane of a surface of revolution is invariant under all rotations about the axis, so it must be horizontal. The unit meridian tangent must consequently have $g^{\prime }\to 0$ and $|f^{\prime }|\to 1$ there. But $|f^{\prime }(\pm \pi /2)|=C$, so smoothness forces $C=1$. The arc-length identity also illustrates the alternatives: $C>1$ prevents reaching either zero with real $g^{\prime }$, while $0<C<1$ gives a nonhorizontal limiting meridian and a conical singularity.

Now $g^{\prime 2}=1-f^{\prime 2}={\cos }^{2}u$. On the connected interior, $g^{\prime }$ has one fixed sign because $\cos u>0$. Therefore $g=g_0\pm \sin u$, andThe revolved surface is the unit sphere centered at $(0,0,g_0)$, with only its two poles $(0,0,g_0\pm 1)$ omitted. Those points are antipodal. This proves the claim using smooth endpoint criterion for a surface of revolution, without integrating a general square root.

The generating circle has $f^{\prime 2}+g^{\prime 2}=1$, and its revolution is a torus. From part (a), $K=\cos u/(2+\cos u)$ and the area element of a surface is $dA=fdudv=(2+\cos u)dudv$. ThereforeThe positive Gaussian curvature on the outer side exactly cancels the negative curvature on the inner side. Equivalently, the Gauss-Bonnet theorem gives ${\int }_{S}KdA=2\pi \chi (T^2)=0$ because the Euler characteristic of the torus is zero.

Write derivatives with respect to $u$. The arc-length condition is $f^{\prime 2}+g^{\prime 2}=1$. The tangent vectors areTheir scalar products give the first fundamental formChoose the unit normal vector $N=(-g^{\prime }\cos v,-g^{\prime }\sin v,f^{\prime })$, since ${\sigma }_u\times {\sigma }_v=fN$ and $f>0$. Scalar products of $N$ with the second partial derivatives giveThe mixed coefficient is zero. Reversing $N$ reverses the second fundamental form but not the Gaussian curvature. Its formula isDifferentiating the arc-length identity gives $f^{\prime }{f}^{\prime \prime }+g^{\prime }{g}^{\prime \prime }=0$. Substitute it directly, without dividing by $g^{\prime }$ (which may vanish), to obtain

Write $F(p)=\mathcal{L}[f](p)$. Assume $f$ and $f^{\prime }$ have right limits at zero and sufficient regularity and exponential-order bounds for the following integration by parts; take $\mathrm{Re}p$ large enough to kill the boundary terms at infinity. First,Apply the same calculation to $f^{\prime }$:This is the Laplace transform of a derivative formula with both initial-data terms retained.

For $a>0$, the Heaviside step function vanishes before $a$ and is one afterwards. For $\mathrm{Re}p>0$,The value assigned at the single jump point does not affect this Laplace transform. The positive real part ensures convergence at infinity.

A quotient topological space of a topological space $X$ is obtained by an equivalence relation $\sim$: its points are the classes $X/\sim$, and the canonical surjection is $q(x)=[x]$. The quotient topology declares a subset $U$ of the quotient open exactly when $q^{-1}(U)$ is open in $X$. More generally a surjection onto a set specifies the same topology by this condition.

For a Hausdorff source with a non-Hausdorff quotient, use $X=\mathbb{R}$ and $x\sim y$ when $x-y\in \mathbb{Q}$. Any nonempty open inverse image is a nonempty open subset of $\mathbb{R}$ invariant under rational translations. It contains an interval $I$, and for every real $t$ there is a rational $r$ with $t-r\in I$; hence it contains every $t$. Thus $\mathbb{R}/\mathbb{Q}$ has the indiscrete topology.

It has distinct points, for instance the classes of $0$ and $\sqrt{2}$, but neither has a proper nonempty open neighbourhood. The quotient is not Hausdorff, even though $\mathbb{R}$ is a Hausdorff space.