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Two counterpropagating modes transform as $(A,B)\mapsto (e^{ik\delta }A,e^{-ik\delta }B)$ under spatial translation. Forcing at twice their natural frequency breaks continuous temporal phase symmetry but preserves simultaneous sign reversal. Hence $\overline{B}$ transforms like $A$ and $\overline{A}$ like $B$, allowing linear parametric coupling. An equal-amplitude phase-locked standing wave has $|A{|}^2=|B{|}^2=z>0$ with $|\beta +\gamma {|}^2{z}^2-2\mu \mathrm{Re}(\beta +\gamma )z+{\mu }^2-{\nu }^2=0$. A nonnegative discriminant is only an algebraic compatibility condition: a physical nonzero wave also requires a positive root.

Subtract the constant equilibrium magnetic energy density. Since ${\mathbf{B}}_0\cdot \mathbf{b}=0$, the perturbation densities areThe traveling-wave amplitude relation from part a givesThis is pointwise Alfvén wave energy transport for one traveling wave, not a claim that every superposition or standing wave has pointwise equipartition.

For the real fields written as $\mathbf{v}=\mathrm{Re}(\hat{\mathbf{v}}{e}^{i(kz-\omega t)})$ and similarly for $\mathbf{b}$, the period averages areThe factors one half distinguish a sinusoid's peak amplitude from its mean squared value; the same complex-vector convention also covers circular polarization of waves.

Use a plane wave proportional to $e^{i(kx-\omega t)}$. Substitution into the convected acoustic wave equation gives the dispersion relationThe two branches have signed phase velocities and group velocities $U_0\pm c_0$. On a branch with $\omega -U_{0}k=s{c}_{0}k$, $s=\pm 1$, the linearized Euler equation givesIn subsonic flow with $0\le U_0<c_0$, the absence of wave reflection selects the outgoing branch on each side:For $U_0=0$, take $A$ real and positive by choosing the source phase, and put $k_0=\omega /c_0$. At $\omega t=2n\pi$, the requested spatial profiles areFor complex $A$, retain $\mathrm{Re}[A{e}^{i{k}_0|x|}]$ instead. The normalized sound pressure and acoustic density perturbation lie on the same even cosine curve. The acoustic particle velocity follows that curve on the right and its negative on the left, as in the figure in part 2(a). Compressed fluid moves away from the source on both sides. These are two outgoing travelling waves, not a standing wave. The ideal point loudspeaker supplies the discontinuity of velocity at $x=0$; the source-free equations apply on either side of it.

Use separation of variables, $y(x,t)=X(x)Q(t)$. The wave equation implies $X^{\prime \prime }/X=\ddot{Q}/(c^{2}Q)=-k^2$, giving $X^{\prime \prime }+k^{2}X=0$ and $\ddot{Q}+c^2{k}^{2}Q=0$. The fixed-end boundary conditions $X(0)=X(l)=0$ selectEach such product is a standing wave. A general released string is a superposition of these normal modes, not necessarily a single separated standing wave. Absence of damping keeps their amplitudes constant.

Seek a separated normal mode $y(x,t)=X(x)Q(t)$. Substitution in the wave equation gives $X{Q}^{\prime \prime }=c^2{X}^{\prime \prime }Q$. For an oscillatory mode choose the separation constant $-{\omega }^2$:The fixed endpoints impose $X(0)=X(l)=0$. In $X=A\sin kx+B\cos kx$, the first condition gives $B=0$, and the second requires $\sin kl=0$. Nonzero displacement therefore requiresEach spatial shape $\sin (n\pi x/l)$ has fixed nodes and an oscillating overall amplitude: a standing wave. A general lossless motion is a superposition of these standing normal modes, not necessarily one single-frequency standing wave. The $n=0$ solution is trivial under both fixed-end conditions and does not supply an additional vibrational mode.

For either normal mode, let $k_n=n\pi /L$ and write its contribution as $B_n\sin (k_{n}x)\sin ({\omega }_{n}t)$. The product-to-sum identity givesThis standing wave is a superposition of two travelling waves of equal amplitude $|B_n|/2$, moving oppositely with phase velocity ${\omega }_n/k_n=c$. Their cancellation at $x=0,L$ enforces the fixed-end boundary conditions at every time.

A standing wave has a fixed spatial amplitude profile and oscillates in time, with stationary nodes of a standing wave. It arises from equal-amplitude waves of the same frequency travelling in opposite directions. For a uniform string with linear mass density $\mu$, the wave speed is $c=\sqrt{T/\mu }$ and a separated mode has $Y^{\prime \prime }+k^{2}Y=0$.

The fixed boundary requires $Y(0)=0$, while the free boundary requires $T{Y}^{\prime }(L)=0$. Thus $Y=A\sin kx$ and $\cos kL=0$, givingThe fixed end is a node of a standing wave and the free end an antinode. Arbitrary small motions are superpositions of these normal modes; the tension alone determines their frequencies only when $\mu$ is also specified.

For two equal-amplitude waves with the same frequency and wavenumber, choose their phases so thatThe spatial amplitude and temporal oscillation separate. Positions with $\cos kx=0$ are fixed nodes of a standing wave, while positions with $|\cos kx|=1$ are antinodes. Different initial phases merely shift this pattern in space and time.

A travelling wave has a phase profile that moves at its phase velocity; a standing wave has stationary nodes and each point oscillates with its fixed local amplitude. The two counterpropagating waves carry equal opposite time-averaged energy flux, so their superposition has zero net time-averaged energy transport. Instantaneous energy can still move within the standing pattern.

For opposite propagation directions on a uniform string, write the equal-frequency transverse waves asThe cosine addition identity givesThis is a standing wave: the spatial pattern is fixed while its amplitude oscillates in time. Nodes occur at the zeros of the first cosine and are separated by $\pi /k$, half a wavelength; antinodes have displacement amplitude $2A$. The two equal opposite travelling waves carry equal opposite average energy flux, so the net average transported power is zero.

For the same propagation direction, the same identity instead givesFor a nonzero resultant this is another travelling wave, with fixed amplitude and moving constant-phase points, not a nontrivial standing wave. If the prefactor vanishes, the waves cancel identically; the zero disturbance is a trivial exception rather than a standing oscillation. The argument assumes propagating waves with $k,\omega >0$ on the same uniform string.

Take the leading real field to be $A{e}^{i(k_{c}x+{\omega }_{0}t)}+B{e}^{i(-k_{c}x+{\omega }_{0}t)}+\text{complex conjugate}$. A spatial translation gives $(A,B)\mapsto (e^{i{k}_c\delta }A,e^{-i{k}_c\delta }B)$; a time translation gives $(A,B)\mapsto (e^{i\psi }A,e^{i\psi }B)$; spatial reflection exchanges $A$ and $B$.

For a monomial $A^j{\overline{A}}^k{B}^l{\overline{B}}^m$ in the first amplitude equation, covariance requires $j-k+l-m=1$ and $j-k-l+m=1$. Hence $j-k=1$, $l-m=0$. At linear order only $A$ is allowed, no quadratic monomial satisfies these conditions, and the cubic possibilities are $A^2\overline{A}$ and $AB\overline{B}$. Reflection fixes the coefficients in the second equation. Removing the small common linear frequency correction by a rotating frame makes $\mu$ real. This derives the two cubic amplitude equations, rather than merely postulating them.

Let $r=|A{|}^2$, $s=|B{|}^2$. Taking real parts givesA travelling wave has $(r,s)=(\mu /{\beta }_r,0)$, or its reflection, whenever $\mu /{\beta }_r>0$. Its phase evolves at $-{\beta }_i\mu /{\beta }_r$, shifting the physical oscillation frequency. The nonzero-amplitude radial eigenvalue is $-2\mu$, while the absent mode has real growth rate $\mu (1-{\gamma }_r/{\beta }_r)$. Therefore, on the supercritical side,The phase direction is neutral because of the symmetry, so stability means orbital stability. A branch with $\mu <0$ is radially unstable, whatever the transverse coefficient.

A standing wave has $r=s=h=\mu /({\beta }_r+{\gamma }_r)>0$. Its two phases evolve together at $-({\beta }_i+{\gamma }_i)h$; the relative phase fixes the spatial position of the standing pattern. In the two intensity directions the eigenvalues are $-2\mu$ and $-2h({\beta }_r-{\gamma }_r)$. ThusAgain the symmetry phases are neutral and the conclusion is orbital stability. Equality ${\beta }_r={\gamma }_r$ leaves a continuum of mixed intensities at cubic order; equality ${\beta }_r+{\gamma }_r=0$ removes cubic standing-wave saturation. These degeneracies require higher-order terms and cannot be classified by the cubic truncation. The zero state is linearly stable for $\mu <0$ and unstable for $\mu >0$.

Write $\psi =\theta +\phi$, $U={\beta }_r+{\gamma }_r$ and $V={\beta }_i+{\gamma }_i$. The polar amplitude equations arewith their reflected counterparts. On the equal-amplitude standing wave branch $R=S>0$, equilibrium point requiresSquaring and adding eliminates the phase:For $U^2+V^2>0$ the candidate squared amplitudes areFor each positive root, set $\sin \psi =-Vz/\nu$ and $\cos \psi =(Uz-\mu )/\nu$. The quadratic identity ensures these define a phase. Conversely these phase equations imply the quadratic, so a nonzero phase-locked branch exists exactly when the discriminant is nonnegative and at least one root is positive.

Since $|\cos \chi |=|V|/\sqrt{U^2+V^2}$, the discriminant condition is the printed $|\mu \cos \chi /\nu |\le 1$ for $\nu \ne 0$. It is sufficient under the usual above-onset supercritical assumptions $\mu >0,U>0$, because then the plus root is positive. Without these assumptions the condition is not sufficient: $\mu =-1,U=1,V=0,\nu =1/2$ satisfies it but gives $z=-1\pm 1/2<0$. This is the positivity condition for a phase-locked standing wave, not a removable algebraic detail. The zero solution always exists and is distinct from the nonzero branch. For $\nu =0$, the undivided equations must be used; a nonzero equilibrium point in this fixed rotating frame requires $V=0$ and $z=\mu /U>0$. If $U=V=0$, nonzero equilibrium points require ${\mu }^2={\nu }^2$ and a compatible phase, with no cubic amplitude selection.

Strong enough forcing can overcome the nonlinear frequency shift, lock the sum phase to the external signal and produce a standing wave at half the forcing frequency. The difference phase remains free through spatial translation. The amplitude inequality is an existence condition, not a stability theorem: attraction requires the linearized forced amplitude equations, and nonlinear coefficients have not been restricted enough here to promise stability of every root.

Once $C$ is removed, the remaining connected segments form the chain $K\to L\to N\to M$, of total length $3\ell$. For $RY\gg 1$, each hydraulic resistance draws a small flow, so to leading order the far end at $M$ is reflecting, with zero segment flow there. Successive wave transfer matrices of an arterial segment giveThe incident-plus-reflected inlet relation remains $q_K=2Y(2P_A-p_K)$. Therefore, defining $D=3e^{3i\beta }+e^{-3i\beta }$,The modulus of $D$ equals $\gamma =|3e^{4i\beta }+e^{-2i\beta }|$, because the expressions differ by a unit-modulus factor. Comparison with the normal flows gives the amplitude ratiosFor small phases the cosines are positive and the absolute values can be removed. The original PDF prints an additional $\cos (\beta /2)$ in the middle numerator. That factor does not follow from the displayed equal-length network: the transfer relation from $M$ to $K$ above fixes $p_M=8P_A/D$, with no such factor. The corrected middle ratio is shown here.

Since ${\gamma }^2=10+6\cos 6\beta$, one has $\gamma =4[1-27{\beta }^2/8+O({\beta }^4)]$. Thus the three corrected ratios, in the order $N,M,L$, areAll three exceed one for sufficiently small nonzero $\beta$; they equal one at zero frequency. This is an oscillatory arterial wave reflection effect, not an increase in steady conductance. Removing a branch changes the standing wave pattern and the total inlet pressure associated with a fixed incident wave. Reflected waves can enhance junction pressures and hence peripheral flow amplitudes. Neither total inlet pressure nor total inlet flow was held fixed, and no extra energy source is implied.

Take the undisturbed free surface as $z=0$, the bottom as $z=-h$, and the walls as $x,y=0,a$. For an incompressible flow which is an irrotational flow, write $\mathbf{u}=\nabla \varphi$ with a velocity potential satisfying ${\nabla }^2\varphi =0$. Impermeability gives ${\varphi }_z=0$ at the bottom, ${\varphi }_x=0$ at the $x$ walls, and ${\varphi }_y=0$ at the $y$ walls.

For the actual surface $z=\eta (x,y,t)$, the exact kinematic boundary condition for a free-surface graph isThe Unsteady Bernoulli equation, with atmospheric pressure at the surface and its time-dependent constant absorbed into $\varphi$, givesFor small disturbances about fluid at rest, $\varphi$ and $\eta$ are first order. Products of first-order quantities and Taylor corrections from evaluating at $z=\eta$ are second order. The linear surface conditions are thereforeor, after eliminating the elevation,

The separated normal modes of surface gravity waves in a rectangular tank havewhere $m,n$ are nonnegative integers not both zero. The horizontal factors satisfy the wall conditions, and the vertical factor satisfies both the Laplace equation and the bottom condition. At the top, ${\varphi }_z=k_{mn}\tanh (k_{mn}h)\varphi$; substitution into the surface condition givesThe elevation isReal physical standing waves are the real parts or linear combinations of these complex modes. The uniform $m=n=0$ mode has zero frequency and would change the mean depth; conservation of the fixed water volume excludes it from the disturbances. Modes with equal $m^2+n^2$ are degenerate and can be superposed.

Rotating Rayleigh-Bénard convection combines buoyancy-driven instability with the Coriolis force. Consider a plane layer of depth $d$ rotating uniformly about the vertical axis, heated from below. Adopt the Boussinesq approximation, fixed boundary temperatures and, for explicit formulas, impermeable stress-free boundary conditions. The conductive state of Rayleigh-Bénard convection is motionless with a linear temperature profile. The dimensionless controls are the Rayleigh number, Prandtl number and Taylor number:Here $\nu$ is kinematic viscosity and $\kappa$ thermal diffusivity. In thermal-diffusion time units, linear perturbations satisfyRotation does no direct mechanical work, since $\mathbf{u}\cdot ({\mathbf{e}}_z\times \mathbf{u})=0$, but couples vertical motion to vertical vorticity and changes the damping and oscillation balance.

For horizontal wavenumber $k$ and vertical mode $n$, put $m=n^2{\pi }^2$ and $a=k^2+m$. Use $w=W\sin (n\pi z)$, $\theta =\Theta \sin (n\pi z)$ and vertical vorticity $\zeta =Z\cos (n\pi z)$ with growth rate $\sigma$. Curling the momentum equation and eliminating pressure givesTheir determinant, without dividing by a possibly zero factor, is the rotating-convection growth-rate polynomialThis makes the linear stability analysis question precise: onset occurs when a root reaches zero real part and all other modes still decay.

A stationary neutral root has $\sigma =0$, givingRotation raises this stationary threshold. An oscillatory neutral root has $\sigma =i\varpi$ with $\varpi \ne 0$. Real-imaginary separation gives the oscillatory neutral curve of rotating convectionThis branch is admissible only when ${\varpi }^2>0$, requiring $P<1$ and sufficiently strong rotation. The restoring Coriolis force coupling permits an inertial/thermal oscillation whose phase-lagged buoyancy can overcome dissipation. At large $P$, temperature and momentum diffusion do not permit that overstability mechanism at primary onset, so the exchange of stabilities is stationary. The actual threshold is the minimum of the stationary and admissible oscillatory curves over all allowed modes, not an arbitrary formal value of $R_o$. Rigid plates require a different vertical eigenproblem and Ekman layers, so the explicit free-slip numbers are not universal.

At large Taylor number, the first vertical mode is selected in the ideal plane layer. Let $t=k^2,m={\pi }^2$. Minimizing the stationary curve givesThus the stationary neutral curve of rotating convection hasThe physical horizontal wavelength is $2\pi d/k_c$, hence decreases as ${\mathrm{Ta}}^{-1/6}$; its prefactor depends on the boundary convention. Thin nearly vertical cells reconcile the strong Coriolis force constraint with viscous and thermal diffusion. The oscillatory minimization replaces the right side of the wavenumber equation by $P^2\mathrm{Ta}m/(1+P{)}^2$, giving the same Taylor number exponent at fixed positive $P$. Where its frequency remains admissible,Equality is $8P^4-P-1=0$, whose positive root is approximately $P_{\ast }=0.6766$. Accordingly, for sufficiently rapid rotation in this free-slip problem, $P<P_{\ast }$ selects oscillatory onset and $P>P_{\ast }$ stationary onset. The weaker condition $P<1$ is only necessary for an oscillatory neutral mode; it does not by itself identify the first instability. Finite Taylor number, finite lateral geometry, allowed discrete wave numbers and plate conditions change the selection.

For the counterpropagating Hopf amplitudes in rotating convection near a simple oscillatory onset, the Hopf bifurcation produces slow complex amplitudes for counterpropagating roll waves. After separating the fast carrier oscillation, symmetry permits the cubic equationswith generally complex coefficients; an $i{\omega }_0{Z}_{\pm }$ term restores the fast frequency if desired. The real parts govern amplitude saturation and the imaginary parts give nonlinear frequency shifts. Write $a_s=\mathrm{Re}{g}_s$, $a_c=\mathrm{Re}{g}_c$. For a travelling wave from a supercritical bifurcation with only one amplitude nonzero, $|Z_{\pm }{|}^2=r/a_s$ with $r,a_s>0$, and the competing wave's growth rate is $r(1-a_c/a_s)$. It is amplitude-stable against that competitor when $a_c>a_s$. A standing wave has equal intensities $r/(a_s+a_c)$; provided this is positive, its intensity-difference mode is stable when $a_s>a_c$. Temporal and spatial phase symmetries leave neutral phase directions, so these are orbital/amplitude stability statements, not decay of every phase displacement.

These coefficients follow from nonlinear interactions and the Fredholm solvability condition obtained by projection onto an adjoint eigenfunction; symmetry alone cannot decide their signs. A negative saturating coefficient gives subcritical bifurcation behavior requiring higher-order terms. Spatial modulation leads to coupled complex Ginzburg–Landau equations with group velocities and diffusion; phase instabilities, mean-flow coupling and differently oriented rolls can destabilize a wave stable in the restricted two-amplitude system. A weakly nonlinear expansion of oscillations therefore predicts travelling waves or standing waves, frequency shifts, modulation and possible secondary mode competition, not a unique universal periodic state.

The Küppers–Lortz instability is a different route to time dependence: it destabilizes steady saturated rolls against oblique roll perturbations. For stationary-roll amplitudes of orientations ${\varphi }_j$, a leading competition system hasA pure roll has $|A_i{|}^2=r/g_0$ with $g_0>0$. An infinitesimal new roll at relative angle $\vartheta$ grows atFor sufficiently strong rotation in appropriate boundary and Prandtl number regimes, some finite oblique angle has $g(\vartheta )<g_0$, so a steady roll is unstable arbitrarily close above its stationary onset. Rotation is handed and allows $g(\vartheta )\ne g(-\vartheta )$, so replacement of one roll by another can favor a definite cyclic sense. Three or more competing orientations can form a heteroclinic cycle; whether it attracts depends on contraction/expansion rates and other modes. Noise, spatially varying domains and modulation can turn this competition into repeated orientation switching and irregular patterns.

The invading rolls are three-dimensional disturbances even when the original straight roll is described by a two-dimensional section. The finite-angle Küppers–Lortz instability mechanism should also be distinguished from the small-angle instability of rotating convection rolls mediated by large-scale mean flow at finite Prandtl number. Numerical thresholds and favored angles depend on mechanical boundaries and material parameters; the essential criterion is the cross-coupling relative to self-saturation. Rotation both changes primary onset and wavelength, and can prevent the resulting steady roll pattern from remaining a stable nonlinear state.

Let $g=a{U}_0\gg 1$ and $z=ka$. A denominator zero away from the removable zero at $z=0$ obeysNear $z=\pi$, write $z=\pi +\delta$ and choose the logarithm branch with $\delta \to 0$. Expanding givesConsequentlyThe lower-half-plane pole is a scattering resonance: an odd standing wave is nearly confined between the two high repulsive barriers but leaks out slowly. If $E=E_R-i\Gamma /2={\hslash }^2{k}^2/(2m)$, then $\Gamma \simeq 2{\hslash }^2\pi \gamma /(m{a}^2)$ and its probability lifetime is $\hslash /\Gamma$. It is not a genuine bound state of the repulsive potential.

The central stratified layer supports upward and downward internal gravity waves, whose counterpropagating components form a standing wave. The unstratified regions support evanescent waves, and the upper disturbance decays as $e^{-k(z-2H)}$. Consequently the central layer is a stratified internal-wave guide, rather than a source of propagating energy at infinity.

The largest response occurs at its trapped normal mode frequencies. They can be specified without solving the forced problem. Set the boundary forcing to zero to find a free normal mode. The lower solution is proportional to $\sinh (kz)$, so continuity of fluid pressure and vertical velocity givesIn the central layer writeThe upper Robin boundary condition then gives the exact trapped-mode conditionThe left side increases strictly from zero to infinity, so there is one positive root for each $n$; the frequencies accumulate at zero. This is constructive phase matching after reflection at both ends. The phase shifts from the evanescent waves matter: simply imposing integer half-wavelengths across the stratified layer is generally incorrect.

In the ideal inviscid model, exact resonant forcing has no bounded steady harmonic solution. The undamped normal mode grows secularly under sustained forcing. Weak viscosity or other losses would produce large finite peaks near the displayed frequencies; the linear approximation eventually fails if the disturbance becomes too large.

The illustrated free normal mode has an oscillatory central region and evanescent wave tails. Its lower tail reaches the fixed zero-displacement boundary; its upper tail decays to infinity.

Choose $F(s)=G(s)=\frac{1}{2}\sin s$. The two travelling profiles then add by the sine addition identity:This is a standing wave. At $x=0$ and $x=\pi$ it vanishes for all $t$. At $t=0$ its displacement is $\sin 2x$, and its time derivative $-5\sin 2x\sin 5t$ is zero. Also $y_{tt}=-25y$ and $y_{xx}=-4y$, directly checking the original equation.

The initial displacement is already the first spatial normal mode, so sine orthogonality gives $A_1=1$ and every other $A_n=0$. Zero initial velocity gives every $B_n=0$. ConsequentlyBoth fixed-end values vanish for all time, and differentiation shows that the initial velocity is zero. This is the fundamental standing wave, with angular frequency $\pi c/L$.

The standing wave oscillates about the constant level $1/2$, with time period $2\pi /|v|$. Suitable snapshots areThe nonconstant amplitude shrinks to zero at a quarter-period, reverses sign at half a period, and returns after a full period. The nodes of a standing wave of the oscillatory part, $x=\pi /2+k\pi$, remain at height $1/2$.

For a plane electromagnetic wave with linear polarization, write $\mathbf{E}=E_0\hat{\mathbf{x}}\cos (kz-\omega t)$ and $\mathbf{H}=E_0\hat{\mathbf{y}}\cos (kz-\omega t)/({\mu }_{0}c)$. The Poynting vector is instantaneous energy flux:The light intensity is its cycle average, not its peak:An incident electromagnetic wave transports longitudinal momentum $\mathcal{E}/c$ per energy $\mathcal{E}$. Perfect reflection back along the incident direction reverses that momentum, transferring $2\mathcal{E}/c$ to the mirror. Dividing by illuminated area and time gives the mean radiation pressureHere $\mathbf{S}$ denotes the incident travelling wave. The average net Poynting vector of the incident-plus-reflected standing wave is zero, although its radiation pressure is nonzero.