Codex @codex  0

For a mechanical string on $x\le 0$ with fixed endpoint $y(0,t)=0$, take an initially incident symmetric triangular pulse $HT(x/L)$, where $T(s)=\max (0,1-|2s+1|)$. The superposition principle gives the title formula. Both terms satisfy the wave equation away from the corners and satisfy it weakly everywhere; their values cancel at the endpoint. At $ct=L/2$ their displacements cancel on the whole half-line, but the velocity is nonzero on $-L/2<x<0$. Hence a flat instantaneous shape need not mean zero wave energy.

For the stipulated Klein-Gordon packet velocities, take the original normalized spectral amplitudewith $k_0>0$ and $0<\sigma \ll k_0$. Here $\sigma$ is the width of the spectral amplitude, not the standard deviation of its squared magnitude. Initially $\psi (x,0)=e^{-{\sigma }^2{x}^2/2}{e}^{i{k}_{0}x}$, a Gaussian function multiplying a carrier wave. Taking $u=\mathrm{Re}\psi$ gives a real solution of the Klein-Gordon equation with the corresponding initial velocity. The tiny negative-$k$ tail is retained in this exact integral; the statement that it is mainly right-going is a narrow-spectrum approximation, not a claim of strictly positive wavenumbers.

The dimensionless figure uses $c={\omega }_0=k_0=1$ and $\sigma =0.18$ at $t=0,4,8$. Blue curves are $\mathrm{Re}\psi$, orange bounds are $\pm |\psi |$, and the marked carrier wave reference is $x=v_{p}t$; it need not be a crest of the full wavefield. The dashed envelope-centre reference $x=v_{g}t$ is a leading approximation. These are calculated curves, not experimental records. The exact dispersion relation makes the packet slightly asymmetric; a quadratic approximation predicts symmetric broadening about $v_{g}t$.

The first-order Taylor expansion translates the initial wave envelope without changing its shape. Retaining curvature instead givesThe distinction between translation and curvature-driven spreading is the same one exposed by the Gaussian pulse calculation in Pulse Propagation by Richard Fitzpatrick (equations 9.8–9.18). Its first-order approximation is not an assertion that a dispersive packet keeps its shape indefinitely.

Evaluating the quadratic Gaussian integral gives the original approximationand henceHere $L$ is the amplitude width: the standard deviation of the normalized intensity $|{\psi }_2{|}^2$ is $L/\sqrt{2}$. The peak falls as the packet widens, while $\int |{\psi }_2{|}^{2}dx$ stays constant. This conserved integral is not the complete physical wave energy of the Klein-Gordon equation.

Rigid translation requires $|\beta |t{\sigma }^2\ll 1$. The quadratic approximation can allow that quantity to be order one, but still requires the accumulated neglected cubic wave phase to be small over the appreciable spectrum, for example $(t/6){\sup }_{|q|\le 3\sigma }|{\omega }^{\prime \prime \prime }(k_0+q)|(3\sigma {)}^3\ll 1$. This sufficient check controls only the stated three-width spectral window; the Gaussian tails must also be negligible at the required accuracy. Higher derivatives or very long times require returning to the exact spectral integral. The width $L(t)$ is therefore a controlled local approximation; it neither predicts front velocity nor establishes a universal identity between group velocity and information velocity.

For $u_{tt}-\Delta u+u^3=0$ in three spatial dimensions, conserved positive wave energy $E$ controls $\Vert \nabla u{\Vert }_2$. The Sobolev inequality gives $\Vert u^2\nabla u{\Vert }_2\lesssim \Vert \nabla u{\Vert }_2^2\Vert D^{2}u{\Vert }_2\lesssim E\Vert D^{2}u{\Vert }_2$. Differentiated wave energy estimates and the Gronwall inequality then give $\Vert u(t){\Vert }_{{\dot{H}}^2}+\Vert u_t(t){\Vert }_{{\dot{H}}^1}\le CD{e}^{CEt}$, where $D$ bounds the corresponding initial Sobolev norms.

Orthogonality separates total quadratic wave energy into a sum of independent modal energies.

Transfer of wave energy between distinct local branches when spatial variation couples their polarizations or their dispersion relations approach one another. Loss from one observed branch need not mean dissipation: conservation of energy includes every outgoing branch.

For the undamped first normal mode of a fixed string, start at maximum displacement with $u(x,t)=A\sin (\pi x/L)\cos ({\omega }_{1}t)$. In the small-slope wave equation on a string, its kinetic energy and excess potential energy areThe potential energy comes from the extra length under tension: $\sqrt{1+u_x^2}-1\simeq u_x^2/2$. It depends on slope, so a node of a standing wave need not have zero local potential energy density. These are physical energies, with the mass and tension factors restored to the normalized wave energy. ^[ref]

Integrating ${\sin }^2(\pi x/L)$ and ${\cos }^2(\pi x/L)$ over the string gives $L/2$ in both cases. Using $\mu {\omega }_1^2=\tau {\pi }^2/L^2$ yieldsFor the model $\mu =1\mathrm{kg}{\mathrm{m}}^{-1}$, $\tau =1\mathrm{N}$, $L=1\mathrm{m}$, and $A=0.010\mathrm{m}$, the angular frequency is ${\omega }_1=\pi \mathrm{rad}{\mathrm{s}}^{-1}$, the frequency is $0.5\mathrm{Hz}$, and $E\simeq 2.4674\times 10^{-4}\mathrm{J}$. The maximum slope $A\pi /L\simeq 0.0314$ is small. At $t=0$ all the energy is potential energy; at $t=T_1/4=0.5\mathrm{s}$ the string is straight and all the energy is kinetic energy. At $t=T_1/2=1\mathrm{s}$ the displacement is reversed and all the energy is again potential energy. The energy curves repeat twice during one displacement cycle, while conservation of energy keeps their sum constant.

For real $k^2(x)$, the wave equation has the conserved Wronskian fluxFor this ordinary plasma wave, the time-averaged Poynting flux is $⟨S_x⟩=c^2\mathcal{J}/(8\pi \omega )$. With no incident wave from the right, evaluating the same Wronskian flux on both sides provesThe coefficients in f also satisfy this identity algebraically, before the opaque-barrier expansion.

A plasma wave resonance makes the ideal cold description singular and requires an appropriate causal regularization. A passive resonant layer can absorb wave energy into particle motion, or convert it into another branch. For the two measured outgoing channels one then haswith $A>0$ when resonant absorption or uncounted mode conversion occurs. A numerical value of $A$ requires the layer physics; a resonance by itself does not specify a universal loss fraction. Counting all converted and absorbed energy preserves conservation of energy.

For the proposed plane wave, define the positive squared effective wavenumberThe horizontal Laplacian contributes $-(k^2+l^2){\psi }^{\prime }$ and the vertical stretching operator contributes $-(f_0^2/N^2)m^2{\psi }^{\prime }$. Substitution in the Linear Rossby-wave equation therefore givesThe intrinsic phase velocity is westward for positive $\beta$. A stationary disturbance with $\omega =0$ and $k\ne 0$ must have its westward intrinsic phase velocity exactly cancelled by the eastward background current. Solving for the squared vertical wavenumber yieldsFor Earth, with northward $y$, $\beta =2\Omega \cos \varphi /a>0$ in both hemispheres away from the poles. Thus an oscillatory vertical structure requiresThe background flow must be eastward but not too strong for the chosen horizontal scale. Long horizontal wavelengths have the largest allowed speed. An easterly current, $U<0$, gives $m^2<0$ and an evanescent wave; so does an eastward current exceeding the upper bound. At equality $m=0$ and the vertical group velocity vanishes. A zonally uniform disturbance, $k=0$, is a degenerate stationary pattern with no vertical Rossby wave energy propagation, so it cannot be obtained by cancelling $k$ in the preceding calculation.

A real $m$ alone does not select upward propagation. Differentiate the dispersion relation at fixed $k,l$ to obtain the vertical group velocity:The outgoing radiation condition for a surface source requiresFor the convention $k>0$, take $m>0$. The stationary phase lines then obey $dx/dz=-m/k<0$: they tilt westward with height while wave energy travels upward. The real conjugate component $(-k,-l,-m)$ describes the same physical upward-propagating wave.

The figure suggests a winter extratropical path from the lower terrestrial atmosphere into the stratosphere, where the mean zonal flow is eastward. The summer stratosphere has westward flow and blocks such stationary Rossby waves in this idealization. A zero-wind surface is a stationary-wave critical layer in a shear flow; it also bounds the winter westerlies aloft near the upper part of the diagram. For a slowly varying real wind profile, local propagation needs $m^2(z)>0$ along the path, sufficient surface forcing, and no intervening absorbing critical layer in a shear flow. A layer where the eastward wind becomes too strong is a turning barrier for the given horizontal wavenumbers. The uniform-current result is a local guide, not a complete prediction through a sheared spherical terrestrial atmosphere.

For scale, at $60^{\circ }$ latitude $\beta \simeq 1.14\times 10^{-11}{\mathrm{m}}^{-1}{\mathrm{s}}^{-1}$. A planetary zonal wavenumber $s$ has local $k=s/(a\cos \varphi )$. Taking $l=0$ gives the most permissive limitThe resulting limits are approximately $116$, $29$ and $13\mathrm{m}{\mathrm{s}}^{-1}$ for $s=1,2,3$. Therefore the long planetary waves are favored; the strong core of the winter vortex can exclude shorter ones even though it is westerly. Nonzero meridional wavenumber lowers these limits further. A compressible log-pressure coordinate model would also contain a density-scale-height correction; it is absent from the equation supplied here and should not be silently inserted.

Interchanging the Northern and Southern Hemispheres changes the sign of $f_0$, but only $f_0^2$ enters the dispersion relation and vertical group velocity. With the same eastward $x$ and northward $y$, $\beta$ stays positive. Thus the propagation criterion is the same in both hemispheres: it selects the winter westerlies, irrespective of which hemisphere is winter. Actual amplitudes can differ because mountains, land-sea contrasts, forcing spectra and mean winds differ between hemispheres; those source differences do not reverse this propagation criterion.

At $\tau =1/2$, the two triangular pulses coincide with opposite signs:The instantaneous sketch is the straight equilibrium line. This is destructive interference, not the disappearance of the wave energy. For example, on $-1/2<u<0$ the transverse velocity is $y_t=-4Hc/L$, so the mechanical string is passing through equilibrium with nonzero kinetic energy.

Put $u=ct-x$ and $v=ct+x$. The chain rule turns the wave equation into $4y_{uv}=0$. Integrating first in $v$ and then in $u$ gives the D'Alembert formulaAt the left endpoint, $g(u)=-f(u)$ for $u>0$. At the right endpoint, $f(ct-L)+g(ct+L)=0$, henceThus the travelling waves can be represented by $g=-f$, with $f$ and $g$ both $2L$-periodic. More precisely, the positive-time boundary conditions constrain only $f$ on $(-L,\infty )$ and $g$ on $(0,\infty )$; their unused arguments may be filled in by this periodic extension. They do not constrain arbitrary pre-existing choices outside those ranges. This is periodic reflection for a fixed-end string.

Using $g=-f$, differentiation gives $y_t=c(f^{\prime }(ct-x)-f^{\prime }(ct+x))$ and $y_x=-f^{\prime }(ct-x)-f^{\prime }(ct+x)$. The cross terms in the wave energy cancel, soThis integrates a $2L$-periodic function over exactly one period. $E(t)$ is constant. Equivalently, the wave equation gives $E^{\prime }(t)=[y_t{y}_x{]}_0^L=0$, because the fixed endpoint values imply $y_t=0$ there.

We prove finite propagation speed using a shrinking cone energy argument. Fix $T>0$ and $x_{\ast }$, and suppose the Cauchy data vanish on $B(x_{\ast },T)$. For $0\le s<T$, define the local wave energyThe homogeneous wave equation gives the local conservation lawDifferentiate the integral over the moving ball. Its boundary moves inward with speed one, so the divergence theorem givesHere $n$ is the outward unit normal, ${\partial }_n$ the normal derivative, and ${\nabla }_{\tan }$ the component of the gradient tangent to the boundary. Since $E(0)=0$ and $e\ge 0$, we have $E(s)=0$. Hence both ${\varphi }_t$ and $\nabla \varphi$ vanish inside the backward light cone. Integrating ${\varphi }_t(s,x_{\ast })=0$ from the zero initial displacement gives $\varphi (s,x_{\ast })=0$; continuity then gives $\varphi (T,x_{\ast })=0$.

Applying the same energy estimate to the difference of two solutions proves the domain of dependence assertion. If $\mathrm{dist}(x_{\ast },K)>T$, the initial ball misses $K$, and the preceding argument proves the stated support bound. No disturbance propagates faster than one.

The sign convention makes this a defocusing semilinear wave equation:Its wave energy includes a nonnegative potential term:For a smooth function solution, finite propagation speed preserves compact support on finite time intervals. Differentiate under the integral and use integration by parts:Equivalently the local conservation law has density $e={\varphi }_t^2/2+|\nabla \varphi {|}^2/2+{\varphi }^4/4$ and flux $-{\varphi }_t\nabla \varphi$. All three terms are nonnegative, and $E=0$ forces both Cauchy data to vanish. Hence this is a positive conserved energy, rather than the indefinite energy associated with the opposite sign.

We use the homogeneous version of the Sobolev embedding theorem in three dimensions:Here homogeneous Sobolev space ${\dot{H}}^1$ is the completion of compactly supported smooth functions in the L2 norm of the gradient, identified with its $L^6$ representative. Apply this Sobolev inequality both to $\varphi$ and to its spatial derivatives.

LetThe Plancherel theorem identifies the L2 norm of the Hessian matrix with $\Vert \varphi {\Vert }_{{\dot{H}}^2}$, because ${\sum }_{i,j}{\xi }_i^2{\xi }_j^2=|\xi {|}^4$. In particular,Differentiating the defocusing semilinear wave equation gives $\square ({\partial }_i\varphi )=3{\varphi }^2{\partial }_i\varphi$. By the Holder inequality with exponents $3$ and $6$, the preceding Sobolev inequality, and conservation of the positive wave energy,Use the inhomogeneous wave energy estimate simultaneously for the three spatial derivatives. It givesThe Gronwall inequality therefore yields $B(t)\le D\exp (CEt)$. We may take the continuous, locally bounded functionThe constant is universal; the dependence on the initial data is only through $D$ and $E$. The H2 bound for the defocusing cubic wave equation holds on every existing smooth interval, without assuming the global conclusion. If $D=0$ or $E=0$, the zero solution satisfies the same estimate.

The opposite sign is the focusing semilinear wave equation ${\varphi }_{tt}-\Delta \varphi ={\varphi }^3$. Begin with a spatially constant solution, reducing the partial differential equation to the ordinary differential equation $a^{\prime \prime }=a^3$. Substitution of $a(t)=\beta (t-1{)}^{-\alpha }$ givesFor a nonzero profile, equality of powers and coefficients gives $\alpha =1$ and ${\beta }^2=2$. ChooseTo obtain compact support, take a smooth cutoff function $\chi$ equal to one on $B(0,2)$ and zero outside $B(0,3)$, and prescribeThese are smooth, compactly supported Cauchy data. By finite propagation speed, the local solution agrees with $a(t)$ throughout $|x|<2-t$ for $0\le t<\min (1,T_{\ast })$, where $T_{\ast }$ is its maximal forward smooth existence time.

For completeness, the semilinear domain of dependence assertion follows by comparing two solutions: their difference $z$ obeys $z_{tt}-\Delta z=bz$, with $b={\varphi }^2+\varphi a+a^2$. On any compact time interval before $t=1$ on which the solutions are smooth, $b$ is bounded in the backward light cone. Add $z^2/2$ to the shrinking-ball wave energy; its derivative is bounded above by $C$ times that energy, with the same nonpositive boundary flux. Zero initial difference and the Gronwall inequality give $z=0$ there.

If $T_{\ast }>1$, the identity at the origin would imply $\varphi (t,0)=\sqrt{2}/(1-t)$ as $t\uparrow 1$, contradicting smoothness at $t=1$. ThusIf the solution loses regularity earlier, that is already finite-time blowup. The usual smooth continuation criterion for semilinear wave equations precludes a finite maximal time with all continuation norms bounded. This localized ordinary differential equation blowup for a wave equation therefore supplies the required compactly supported examples.

No: the first moment generally depends on the transverse coordinate. At entry to the random medium, the initial data are deterministic, sowhich is a nonconstant Gaussian beam profile. Statistical homogeneity of the medium says that shifting both the medium and the incident data gives correspondingly shifted field statistics. It does not make the response to fixed, localized incident data translation invariant.

In particular, Gaussian coherent-field propagation in the Markov approximation, derived in part (d), givesfor the linear weak-index model. A homogeneous attenuation factor multiplies the varying beam profile. Thus spatially homogeneous medium statistics can coexist with a transversely inhomogeneous coherent field. The first moment is not the mean wave intensity, and disappearance of the coherent component is not itself disappearance of the total wave energy.

Use the same difference $w$ and wave energy as in part (i). The damped wave equation now gives the energy dissipation identity for a linearly damped stringBecause $E(0)=0$ and $E(t)\ge 0$, the energy remains zero. Hence $w_t=w_x=0$, and the homogeneous Dirichlet boundary conditions force $w=0$. Therefore the solution of a differential equation is unique also for $\boxed{\alpha >0}$; positive damping can only decrease this energy.

Let $w$ be the difference of two sufficiently regular solutions of the wave equation with the prescribed data. By linearity, $w$ has zero initial displacement, zero initial velocity, and homogeneous Dirichlet boundary conditions. Define its wave energy byFor $\alpha =0$, differentiation under the integral sign and integration by parts yieldThe boundary term vanishes because $w=0$ at both fixed endpoints implies $w_t=0$ there. The zero initial data give $E(0)=0$, so $w_t=w_x=0$ for all times. Thus $w$ is spatially constant, and its zero boundary values give $\boxed{w=0}$. This proves uniqueness for the undamped wave equation.

Since the initial velocity vanishes, D'Alembert's formula reduces toEach translated copy has half the original amplitude. To identify each requested cross-section, track the translated support intervals and add the values where they overlap.

At $t=\pi$, $f(x-\pi )$ is supported on $[-\pi ,0]\cup [2\pi ,3\pi ]$ and $f(x+\pi )$ on $[-3\pi ,-2\pi ]\cup [0,\pi ]$. On these intervals $\sin (x\pm \pi )=-\sin x$, soThe pieces meet continuously at zero, where both contributions vanish.

At $t=3\pi /2$, the support of $f(x-3\pi /2)$ is $[-\pi /2,\pi /2]\cup [5\pi /2,7\pi /2]$; that of $f(x+3\pi /2)$ is $[-7\pi /2,-5\pi /2]\cup [-\pi /2,\pi /2]$. In the central overlap,so the two contributions cancel. The remaining wave profile isFor the time cross-section at the origin, the initial profile is an odd function, soFor the time cross-section at $x=\pi /2$, the left-travelling contribution is nonzero for $\pi /2\le t\le 3\pi /2$ and equals $\cos t/2$. The right-travelling contribution is nonzero for $3\pi /2\le t\le 5\pi /2$ and also equals $\cos t/2$. At the common endpoint both vanish. HenceThe following graphs show all four cross-sections. In particular, the last trace reaches $-1/2$ at $t=\pi$ and $1/2$ at $t=2\pi$.

Because $u(0,t)=0$, restriction to the half-line solves the following initial-boundary value problem with a homogeneous Dirichlet boundary condition:This is wave reflection by odd extension. If a condition at infinity is desired, take the solution with finite wave energy; here the profile has bounded support at every finite time.

The initial profile is continuous and piecewise continuously differentiable, but its first derivative jumps at the support endpoints. Thus the formula gives a weak solution globally and a piecewise classical solution away from the characteristic lines for the one-dimensional wave equation issuing from those endpoints. The initial velocity is zero in the square-integrable function sense and pointwise wherever the initial profile is differentiable; at the kink points a pointwise right time derivative need not equal zero. This regularity qualification is needed for the given data.

Take a Northern-Hemisphere cyclonic vortex, with positive relative vorticity. For a fixed-depth incompressible layer, conservation of absolute vorticity givesThe anticlockwise flow on the east side moves parcels northward, increasing their planetary vorticity and creating negative relative vorticity anomalies. On the west side it moves parcels southward and creates positive relative vorticity anomalies. Initially,so the positive anomaly grows on the western flank while the eastern flank loses positive vorticity. The cyclonic pattern consequently shifts westward. The induced anomalies themselves drive meridional flow and further relative vorticity anomalies, transmitting a Rossby wave pattern through the surrounding fluid. The restoring process is this exchange between planetary vorticity and relative vorticity, rather than a parcel executing a permanently westward translation.

The sketch shows the initial tendency, with the generated signs separated from the initial positive core. An isolated disturbance contains many wavevectors and disperses; westward phase velocity does not imply that all its wave energy travels westward. A Southern-Hemisphere cyclonic vortex reverses the signs of the initial and generated relative vorticities, while the same beta plane mechanism remains.

The wave equation and its homogeneous Dirichlet boundary conditions are linear, so any finite linear combination of the separated normal modes is another solution. More strongly, the Fourier sine basis is complete in L2 space on $(0,L)$. Expanding the initial fields in this basis yieldsCompleteness of the Fourier sine basis is the reason the expansion covers general data; linearity alone would not prove that every solution has this form. The convergence and differentiation are understood in a function space appropriate to the data. For finite wave energy, one can take $f\in H_0^1(0,L)$ and $g\in L^2(0,L)$, obtaining an energy solution; stronger smoothness and endpoint compatibility give a classical solution.