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For the scalar conservation law, put $v_0(\xi )=f^{\prime }(u_0(\xi ))$ and $m={\min }_{\xi }{v}_0^{\prime }(\xi )={\min }_{\xi }{f}^{\prime \prime }(u_0(\xi ))u_0^{\prime }(\xi )$. The concave-flux characteristic lifespan and characteristic flow map areFor $t<t^{\ast }$, $X_t^{\prime }=1+t{v}_0^{\prime }>0$, and $X_t(\xi )=\xi +ct$ outside the support of $u_0$. Thus $X_t$ is a global smooth diffeomorphism and the formula is $u(t,x)=u_0(X_t^{-1}(x))$. The method of characteristics proves existence and uniqueness among smooth solutions.

If $m<0$, at a minimizer ${\xi }_{\ast }$ the numerator $u_0^{\prime }({\xi }_{\ast })$ is nonzero, since its product with $f^{\prime \prime }(u_0({\xi }_{\ast }))$ is negative. Thereforeblows up as $t\uparrow t^{\ast }$, showing that this is the maximal smooth lifespan. Strict concavity alone does not require $f^{\prime \prime }$ to be negative at every point; the argument uses no such extra hypothesis.

The Inviscid Burgers equation here has the negative quadratic flux $F(f)=-f^2/2$. By the method of characteristics, $df/dZ=0$ and $d\theta /dZ=-f$, so a characteristic beginning at ${\theta }_0$ satisfiesThis is a classical solution until characteristic curves intersect. Across a shock wave, the Rankine-Hugoniot condition givesIt uses the one-sided limiting values and agrees with the weak-shock expression.

For $U>0$, the right-state characteristic curves travel left with speed $-U$ while the left-state characteristic curves have speed zero. They converge into a compressive shock wave with speed $-U/2$. The entropy solution isThe characteristic speeds satisfy $0>-U/2>-U$, so both sides enter the shock.

For $U<0$, right-state characteristic curves travel right with speed $-U>0$, leaving a gap. A steep continuous transition fills that gap with a rarefaction wave. The entropy solution isThe middle value is fixed by $\theta /Z=F^{\prime }(f)=-f$. A discontinuity joining the two states would satisfy the jump equation but violate entropy admissibility. If $U=0$, the solution is identically zero.

Let $s$ be arclength, $\mathbf{t}={\partial }_s\mathbf{r}$ the unit tangent, and $\mathbf{v}$ the local filament velocity relative to the fluid. The resistive-force theory force per unit arclength, exerted by the fluid on the filament, isThe parallel and perpendicular drag coefficients of a slender filament are proportional to dynamic viscosity. For radius $a$ and a relevant axial cutoff length $\ell \gg a$, the anisotropic slender-filament drag from Stokeslet integration gives the leading logarithmic formsThe cutoff may be set by filament length or deformation wavelength. End shape and the choice of cutoff change order-one constants inside the logarithms. Slender-body theory also includes nonlocal interactions that the local resistive-force theory approximation omits. Typically ${\xi }_{\perp }>{\xi }_{\parallel }$, with ratio tending to two in the slender limit; this drag anisotropy enables propulsion.

Write $C={\xi }_{\perp }-{\xi }_{\parallel }>0$. For a small-amplitude planar deformation, $\mathbf{t}=(1,y_x)+O(y_x^2)$ and the leading transverse velocity is $y_t$. Exact inextensibility requires the quadratic longitudinal displacement of an inextensible planar filament: if $X(s,t)=s+X_2(s,t)$ and $Y(s,t)=y(s,t)$, then $X_{2,s}=-y_s^2/2$ to this order. Its velocity is quadratic and has zero time average for a periodic material cycle without mean axial drift. Thus the fixed material abscissa in the question is leading-order notation; this correction does not change the averaged propulsive term below. An imposed mean translation would contribute its separate ordinary axial drag.

To leading nonzero order,The time-averaged deformation-induced propulsive force of a periodic planar filament and power of a periodic planar filament are consequentlyThese are the leading quadratic functionals. Their signs refer to force on the filament. For example, $y=a\sin (kx-\Omega t)$ gives negative $F_x$, opposing the direction of wave propagation.

Let $\eta (x,t)$ be an arbitrary smooth variation periodic in both variables. The first variation of the constrained functional isUsing integration by parts in $x$ and $t$, every boundary term cancels by periodic boundary conditions. ThereforeThe Euler-Lagrange equation isFor a nontrivial stroke with nonzero thrust, $\Gamma \ne 0$. Set $c=C/(\Gamma {\xi }_{\perp })$ and $v=y_t$. Then $v_t+c{v}_x=0$, whose method of characteristics solution is $v=V(x-ct)$. Integrating in time gives the travelling-wave stationarity of planar filament propulsion:Here $g$ is an arbitrary static periodic component. It contributes neither power nor mean propulsion because the time average of $y_t$ vanishes. It can be set to zero when describing the active stroke. The wave must be compatible with both specified periods: $f(x-cT)=f(x)$.

For a pure travelling wave, its leading force and power areThus positive $x$ propulsion uses a wave travelling in the negative $x$ direction. Any smooth interior maximizer satisfying only the stated power constraint must have this travelling active deformation. If the drag were isotropic, $C=0$ and there would be no leading propulsion.

There is a qualification to the printed maximization claim: the variation proves stationarity, not the existence of an unrestricted global maximum. The bandwidth limitation in fixed-power filament optimization can be seen directly. Set $k=2\pi /\lambda$, $\Omega =2\pi /T$, and consider $y_m=a\sin (mkx+\Omega t)$. The leading expressions giveAt fixed power the formal quadratic objective increases with $m$. Eventually that sequence violates the small-slope approximation, but the approximation supplies no numerical slope or wavelength cutoff with which to define the global optimization. Even within a strict small-slope neighbourhood, a sufficiently small admixture of a higher spatial harmonic can improve a candidate interior maximum while keeping power fixed. For example, $y=a[\sqrt{1-{\eta }^2}\sin (kx+\Omega t)+\eta \sin (mkx+\Omega t)]$ retains the fundamental periods and exactly the same leading power, while its force is multiplied by $1+(m-1){\eta }^2$. For fixed $m>1$, taking $a$ and $\eta$ small keeps every slope small.

With an explicit admissible Fourier bandwidth or an additional geometric constraint, the intended conclusion can be made precise. For a finite set of nonstatic Fourier series modes proportional to $e^{i(k_{j}x-{\Omega }_{j}t)}$, with wavenumbers $k_j$ and frequencies ${\Omega }_j$,This is a weighted average of $-C{k}_j/({\xi }_{\perp }{\Omega }_j)$. A largest allowed ratio is attained by a single travelling mode, or by modes sharing the same phase speed; their superposition is still a travelling wave. Thus the Euler-Lagrange calculation yields the requested travelling-wave form, while a global maximum requires a specified admissible shape class.

Assume $\omega \ne 0$. With $h=0$, the method of characteristics gives $f(t,\Phi (t,0,z))=f_0(z)$. HenceLet $K=\mathrm{supp}{f}_0$ and $E_{\ast }={\max }_K{H}_{\omega }$; if $f_0=0$, the conclusion is immediate. By the Hamiltonian energy balance, every image point of $K$ remains in the energy sublevelThis is a fixed compact ellipsoid in phase space. Therefore the support bound is uniform in time:This uniform support bound from a coercive conserved energy requires no explicit solution of the curves.

The nonzero-frequency qualification is necessary. At $\omega =0$ the energy does not control $x$, and the equation is free transport equation. Choose a smooth compactly supported $f_0$ that is nonzero at $(x_0,v_0)$ with $v_0\ne 0$. Its transported value at $(x_0+t{v}_0,v_0)$ stays nonzero, so the union of the supports is unbounded. Each individual support is compact, but there is no fixed compact support for all times.

The first-order Cauchy-Kovalevskaya theorem says that an analytic PDE system solved for the derivative normal to an analytic noncharacteristic initial hypersurface, with analytic initial data, has a unique local analytic solution near each point of that hypersurface. The general theorem likewise applies to higher-order equations solved for the highest normal derivative with analytic data for the lower normal derivatives.

Here the characteristic vector is $(x,y,a)$ and the normal to $z=0$ is $(0,0,1)$. Its normal component is $a$, so the theorem applies precisely when $a\ne 0$. The method of characteristics gives $\dot{x}=x$, $\dot{y}=y$, $\dot{z}=a$, $\dot{u}=u$. Starting from $(\xi ,\eta ,0)$ at parameter $s=0$ gives $x=\xi e^s$, $y=\eta e^s$, $z=as$, $u=e^{s}f(\xi ,\eta )$. Eliminating $s=z/a$ yieldsThis is locally defined wherever the arguments remain in the analytic domain of $f$, and verifies both PDE and initial data.

As $a\to 0$, the normal component vanishes and the characteristic curves become tangent to the initial plane. The crossing parameter $s=z/a$ and slopes $dx/dz=x/a$, $dy/dz=y/a$ become singular. At $a=0$, arbitrary analytic data are not admissible: already on the plane they must obey $x{f}_x+y{f}_y=f$. Even compatible data do not determine derivatives normal to the plane. This is failure of the noncharacteristic condition, not a failure of the analytic theorem.

For the method of characteristics, parameterize by $x$; the characteristic curves satisfyThus $x^2+y=C$ and $u-\sin x$ is constant on each characteristic curve. The general form in the quadrant is $u=\sin x+F(x^2+y)$. The boundary condition gives $F(s^2)=\cos s-\sin s$ for every $s\ge 0$.

The characteristic curve through $(x,y)$ meets the boundary at $(s,0)$ with $s=\sqrt{x^2+y}\ge x$. Integrating backwards from that boundary value yieldsIndeed ${\partial }_x-2x{\partial }_y$ annihilates $x^2+y$ and sends $\sin x$ to $\cos x$, while on $y=0$ the expression reduces to $\cos x$. Every characteristic in the quadrant meets the prescribed boundary, so its value is determined. The displayed solution is smooth away from the origin. At the origin the boundary is characteristic and the solution is continuous but has a singular inward $y$ derivative; the differential equation is interpreted in the interior, with the boundary data attained continuously.

The vertical component $q_z=\zeta -f\eta /H$ is $H$ times the first-order anomaly in shallow-water potential vorticity about rest:Thus $\mathbf{q}=q_z\hat{\mathbf{z}}$ records the combination of relative vorticity and column stretching that is conserved by the unforced linearized shallow water equations. Indeed their curl and continuity equation give $(q_z{)}_t=0$. It is the persistent forcing of the balanced component during geostrophic adjustment, while inertia-gravity waves carry away the unbalanced component.

Initially $q_z=-f{\eta }_0\mathrm{sgn}(x)/H$. In the final geostrophic flow, $u=0$, $v=g{\eta }_x/f$, and $\zeta =g{\eta }_{xx}/f$. Conservation of the linearized shallow-water potential vorticity therefore givesBoundedness on both sides and continuity of $\eta ,{\eta }_x$ at the origin determine the constants:The solution is the locally attained balanced state after outgoing inertia-gravity waves have departed; the full inviscid system need not lose its globally conserved wave energy. The signed volume flux in the $y$ direction is

For double Kelvin-wave adjustment at a depth step, put $c_{\pm }=\sqrt{g{H}^{\pm }}$ and use $R_D^{\pm }=c_{\pm }/f$ on the corresponding side of the step. The PDF's far-field wording must mean $\eta \to {\eta }_0\mathrm{sgn}(x)$, rather than ${\eta }_0$ for every $x$. In the exponentially trapped outer response, cross-step geostrophic balance gives, at $y=0$,To retain the propagation of this slowly varying trapped disturbance, the along-step acceleration in $u_t-fv=-g{\eta }_x$ must also be retained. It supplies the small cross-step ageostrophic flow. Thus the two cross-step volume fluxes areHere $d\mathrm{sgn}(x)/dx=2\delta (x)$ as a distributional derivative. Equating the fluxes and dividing by $c_{+}+c_{-}$ givesThe printed forcing is missing a factor of two. It cannot follow from the printed sign-function surface profile with the usual Dirac delta distribution normalization. Pure geostrophic balance in both horizontal directions would also omit $A_t$ and could not determine the wave evolution.

The method of characteristics with zero initial amplitude gives the corrected weak solutionBehind the eastward front, $0<x<\Delta ct$, the surface elevation at the step becomes $-{\eta }_0$, matching the upstream side. All the original cross-step transport is diverted into flow along the step once the wave has passed a fixed location: $v=0$ there, with opposing along-step velocities above and below the step. This is the complete barrier in the adjusted region, and in the long-time limit at every fixed $x$. At finite time a front still carries cross-step flux: the formulas above give $H^{\pm }{v}^{\pm }=(2{\eta }_0{c}_{+}{c}_{-}/f)\delta (x-\Delta ct)$. The discontinuous outer representation does not resolve the short-scale fast adjustment at either front.

For completeness, solving the equation literally as printed instead gives $A={\eta }_0[\Theta (x)-\Theta (x-\Delta ct)]$. Its amplitude is half that required; $\eta (x,0,t)$ retains a jump ${\eta }_0$ at the origin after passage, so that literal equation does not imply complete blocking. The corrected equation and the literal printed equation are distinguished rather than asserting an inconsistent barrier result.

For real initial data, separation of variables givesThis includes $z_0=0$, whose solution is identically zero. Every other real $z_0$ has a pole at $t=1/z_0$, so only $z_0=0$ gives a finite solution on the whole real line. If complex-valued initial data are allowed, a nonreal $z_0$ has no real-time pole; the whole-real-line conclusion here uses the usual real-valued differential-equation convention.

Along a characteristic curve $y=u+ax$, the chain rule gives $df(x,u+ax)/dx=f_x+a{f}_y$. For the homogeneous transport equation this is zero, so $f(x,y)=F(y-ax)$. For the nonlinear equation, the same method of characteristics reduces the problem to $df/dx=f^2$, with initial value $g(u)$. ThereforeFor differentiable real $g$, this is the classical solution wherever its denominator is nonzero, on the characteristic intervals containing the initial line. If $g(u)\ne 0$ for some $u$, that characteristic blows up at $x=1/g(u)$, $y=u+a/g(u)$. Thus the only real-valued solution bounded on all of ${\mathbb{R}}^2$ has $g\equiv 0$, giving $f\equiv 0$.

The substitution removes the zero-order term: $u_x=e^{-x^2}(v_x-2xv)$ and $u_y=e^{-x^2}{v}_y$, so the partial differential equation becomes $v_x+x{v}_y=1$. Its method of characteristics givesThe invariant is $\eta =y-x^2/2$. Therefore $v=x+F(\eta )$. The initial line $x=0$ is noncharacteristic, and the boundary data give $F(\eta )=\eta e^{-{\eta }^2}$. HenceEach characteristic meets the data line once, ensuring uniqueness among differentiable solutions, and substituting the result verifies both the equation and the data.

Write $H_{\pm }=H_0\pm d$, with the upper sign referring to $y>0$, and use an f-plane with Coriolis parameter $f_0>0$ for the stated directions. The formulas retain the signed $f_0$; a positive Rossby deformation radius means $R_{\pm }=\sqrt{g{H}_{\pm }}/|f_0|$. Let $\rho$ be the constant fluid density.

Far from the depth step the problem is independent of $y$. The linearized shallow water equations imply conservation of $v_x-f_0\eta /H_{\pm }$. Initially this is $-f_0{\eta }_0\mathrm{sgn}(x)/H_{\pm }$. In geostrophic balance, $u_s=0$ and $v_s=g{\eta }_{s,x}/f_0$, soBoundedness and matching of height and its derivative at $x=0$ give geostrophic adjustment of a surface-height jump:The potential energy release per unit length in $y$ is the finite difference between two individually infinite energies:For comparison, the final kinetic energy is $\rho g{\eta }_0^2{R}_{\pm }/2$, so the energy radiated in inertia-gravity waves is $\rho g{\eta }_0^2{R}_{\pm }$. Integrating the depth-weighted velocity over $x$ gives the volume transportsFor $f_0,{\eta }_0>0$, $Q_{-}$ is towards the step and $Q_{+}$ is away from it. Their difference is $4g{\eta }_{0}d/f_0$; a global steady state cannot simply join the two far-field currents without additional transport along the step.

To obtain the rotating shallow-water height equation with variable depth, set $\zeta =v_x-u_y$ and $J_H={\partial }_x(Hv)-{\partial }_y(Hu)$. Direct differentiation of the original momentum and continuity equations givesConsequentlyAt the step, $H_y=2d\delta (y)$ is a Dirac delta distribution; the original matching conditions are used rather than assuming a smooth depth there. Velocity elimination raises the time order: initially ${\eta }_t=0$ and ${\eta }_{tt}=g{\nabla }_h\cdot (H{\nabla }_h\eta )$, as well as the prescribed initial height.

For a decaying harmonic disturbance, the height equation on each constant-depth side givesSolving the two momentum equations yieldsContinuity of $H\hat{v}$ at the step gives the implicit step-trapped topographic Rossby wave dispersion relationThis derivation assumes ${\omega }^2\ne f_0^2$ and positive decay rates; it describes the low-frequency trapped branch, rather than the radiating inertia-gravity waves.

In the quasi-geostrophic approximation, replace $f_0^2-{\omega }^2$ by $f_0^2$. The resulting explicit frequency, and then its leading small-$d/H_0$ form, areFor $d>0$, the phase velocity has the sign of $f_0$. Thus propagation is eastward in the Northern Hemisphere, with the shallower side on the right. Differentiating the leading dispersion relation givesIn the long-wave limit the group velocity and phase velocity coincide:

To find the slow amplitude, integrate the low-frequency height equation across the step. This gives ${\partial }_t[H{\eta }_y]=f_0[H]{\eta }_x$, where brackets mean the upper-side minus lower-side value. For the given exponentially trapped outer profile, $[H{\eta }_y]=2H_{0}A/R_D$ and ${\eta }_x(x,0)=2{\eta }_0\delta (x)-A_x$. Hence long-wave transport along a depth step obeysThe method of characteristics, interpreted for the Dirac delta distribution source, givesFor $c_T>0$ this is $2{\eta }_0$ in $0<x<c_{T}t$ and zero outside: a surface depression is left in the wake of the wave. The sharp $x$-jump is a long-wave outer approximation. The fast adjustment region of width $O(R_D)$ near the origin, and the detailed wavefront, are unresolved. Keeping a smoothed background ${\eta }_s(x)$ instead would give $A_t+c_T{A}_x=c_T{\eta }_s^{\prime }(x)$ and $A={\eta }_s(x)-{\eta }_s(x-c_{T}t)$ at this order.

The along-step geostrophic flow is $u=-g{\eta }_y/f_0$. Integrating separately over the two sides givesIn the wake, $T_{-}=Q_{-}$ and $T_{+}=-Q_{+}$. The shallow-side transport is eastward and the deep-side transport westward, with magnitudes equal to the respective far-field currents. The net along-step transport is $T_{-}+T_{+}=-4g{\eta }_{0}d/f_0$, balancing their mismatch. These comparisons use the stated leading long-wave and small-depth-contrast approximations.

The Collisionless Boltzmann equation is $df/d\eta =0$. At first order, directional deflections act only on the isotropic background and hence give no angular contribution. Differentiating $f=\bar{f}-\Theta d\bar{f}/d\ln ϵ$ along the unperturbed trajectory givesUsing the tensor redshift of the comoving energy,The method of characteristics follows the backward ray $\mathbf{x}({\eta }^{\prime })=\mathbf{x}-(\eta -{\eta }^{\prime })\mathbf{e}$. With zero initial perturbation, the line-of-sight solution isTo this order the frame and coordinate direction components may be identified inside a quantity already proportional to $h$.

The method of characteristics shows that a point is affected only after a westward characteristic from the forcing region reaches it. At a western observation point $X_1$, during the intervalthe fast barotropic signal has arrived but the first and higher baroclinic signals have not. Consequentlywhich is nearly independent of depth. At the eastern point $X_2$, no westward Rossby-wave characteristic arrives from the forcing region, so the disturbance remains zero under the stated initial and radiation conditions.

The propagation speed derived in part v is $C_n=\beta c_n^2/f_0^2$. Thus the question's notation $c_{0}t>L_X>c_{1}t$ must be read as a comparison with these modal Rossby signal speeds; using the gravity-wave speeds $c_n$ literally would not describe the long-wave equation derived above.

The method of characteristics givesAlong a characteristic,Thus the characteristics are circles $x^2+y^2=C$, traversed counterclockwise, and $u$ is constant on each circle. Their intersections with the $x$-axis have $x^2=C$, where the boundary value is $f(C)$. HenceThe characteristic sketch is the family of concentric circles centred at the origin.

For constant $v_m$, $M_r$ obeys the linear transport equationThe method of characteristics therefore givesThe localized profile translates inward without changing shape: material initially concentrated near $r_0$ reaches the central star after

Each annulus loses specific angular momentum as its characteristic moves to smaller $r$. The negative magnetic torque transmits that angular momentum along the large-scale field to the anchored interstellar medium. Thus the disk's angular momentum decreases even before mass crosses $r=0$; angular momentum is conserved only after the field and its anchoring medium are included.

For capillary residual trapping of an initial triangular current, the method of characteristics translates its leading and trailing faces at $v_A=u/\varphi$ and $v_R=u/[\varphi (1-s)]$. Their intersection has height $a[L-(v_R-v_A)t/2]$. Extinction occurs at $t_{\ast }=2L/(v_R-v_A)$ and position $x_{\ast }=2L/s$. These formulas require $0<s<1$; at zero residual saturation the triangle translates indefinitely.