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If the first two phase derivatives vanish at an endpoint but the third does not, an oscillatory contribution is concentrated on a width proportional to $N^{-1/3}$. For a smooth nonzero endpoint oscillatory integral amplitude, rescaling reduces the leading contribution to a cubic oscillatory integral. Its constant follows from the Gamma function Fourier integral. This is a degenerate form of the stationary phase method, and a quadratic formula is insufficient.

For fixed $0<\alpha <\pi /2$, the Bessel function of the first kind has an interior nondegenerate stationary point in its cosine integral. The stationary phase method gives the displayed oscillatory leading term and an additive $O(n^{-3/2})$ remainder. The approximation loses uniformity near $\alpha =0$, where the stationary point reaches the endpoint and the curvature vanishes. For negative integer-order arguments, parity supplies a factor $(-1{)}^n$.

Write $N=\sqrt{\alpha g}$ and $b_n^2=(n\pi /h{)}^2+{\alpha }^2/4$. For the signed rightgoing dispersion relation $\omega (k)=Nk/\sqrt{k^2+b_n^2}$, the horizontal phase velocity is $N/\sqrt{k^2+b_n^2}$ and the group velocity is $N{b}_n^2/(k^2+b_n^2{)}^{3/2}$. Its maximum is $N/b_n$ at zero wavenumber; the leftgoing branch has the opposite speed. The stationary phase method for an initially localized wave packet in this vertical mode gives appreciable large-time contributions on rays satisfying $|x|/t\le N/b_n$. This is an asymptotic propagation region, not exact compact support. Data with algebraic tails retain smaller contributions outside it, and the maximal-speed front needs a degenerate stationary-phase approximation.

Rotating the integral defining the Gamma function toward the positive imaginary axis introduces the factor $i{e}^{-\pi n/2}$. After scaling by $n$, the remaining phase is $\log s-s$, whose unique stationary point is $s=1$ with second derivative $-1$. The stationary phase method gives $e^{-in-i\pi /4}\sqrt{2\pi /n}$ for this scaled integral, producing the displayed asymptotic. The boundary contour integral must be Abel-regularized; its off-saddle tails are smaller by integration by parts.

In an integral $\int a(q)e^{i\lambda \beta (q)}dq$, the real-valued function $\beta$ is the phase. Its stationary points satisfy $\nabla \beta =0$; its Hessian matrix determines the local quadratic oscillation and the signature factor in stationary phase method. Degenerate stationary points and boundaries require separate analysis.

A chirp is variation of the instantaneous frequency through a signal. Here the group delay relative to the pulse centre at the optical lens isConsequently $\eta >0$ orders lower frequencies before higher frequencies; $\eta <0$ reverses that ordering. A representative strongly chirped optical pulse iswith a smooth localized envelope $A(t)$. The carrier oscillations become closer together towards later times for positive $\eta$ and farther apart for negative $\eta$.

The relation between the spectral phase and a locally assigned instantaneous frequency is a stationary phase method, not an exact reconstruction from spectral phase alone: the spectral wave amplitude also matters. For example, if the spectral envelope is $e^{-(\omega -{\omega }_0{)}^2/(2{\sigma }^2)}$, a Fourier transform of a Gaussian gives the exact real-field phase ${\omega }_{0}t+\eta t^2/(4|a{|}^2)$ up to a constant, where $a=1/(2{\sigma }^2)-i\eta$. The slope retains the sign of $\eta$ and approaches $1/(2\eta )$ for $|\eta |{\sigma }^2\gg 1$.

For asymptotic analysis it is useful to separate the spectral density from the observation time. For a fixed horizon $T\ge t$, $H(k,T)$ already has this property. If the boundary traces extend to all positive times with suitable growth, their one-sided Laplace transforms give the infinite-horizon versionFor bounded traces these integrals converge in $D_{+}$; boundary values are taken by an Abel limit where needed. The finite-horizon formula permits this replacement: the tail from $t$ to infinity contributes an analytic, decaying integrand with zero integral around $\partial D_{+}$. The resulting spectral functions are independent of $t$.

They can be made particularly explicit on the two contour pieces. On the real axis and upper curve, respectively,The second identity follows because $k^3+k$ is real on $C$: the remaining roots are $\overline{k}$ and $-2\mathrm{Re}k$. Therefore rewrite the solution asThis isolates all large-time dependence in the exponential, allowing the stationary phase method or the method of steepest descent. Along $x/t=\xi$, the phase is $i[k\xi -k-k^3]$, and its stationary points satisfy $3k^2=\xi -1$. Thus the real saddles coalesce at $\xi =1$; any poles arising from the boundary Laplace transforms also have to be retained during contour deformation.

The source specifies smooth boundary traces, not their large-time decay or growth. Accordingly a universal numerical long-time profile cannot be inferred. The infinite-horizon formula requires the stated additional transform hypothesis; for more general data the finite-horizon density $H(k,T)$ remains valid, or an appropriately shifted Laplace contour must be used. This qualification does not leave an unknown boundary trace in the representation.

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 $U$.

Use an acoustic velocity potential $\phi$ with $\mathbf{u}=\nabla \phi$ and $p^{\prime }=-{\rho }_0{\phi }_t$. It obeys the wave equation ${\phi }_{tt}=c_0^2{\nabla }^2\phi$. The rigid walls require zero normal velocity, hence Neumann boundary conditions on the transverse variables. Separation givesDefine ${\kappa }_{mn}^2=m^2/{\alpha }^2+n^2/{\beta }^2$. The dispersion relation, phase velocity and group velocity for positive $k,\omega$ areFor a dispersive mode, $\omega (0)=c_0{\kappa }_{mn}>0$, $c$ decreases from infinity to $c_0$, and $c_g$ increases from zero to $c_0$. The $(0,0)$ mode has $\omega =c_{0}k$ and $c=c_g=c_0$: it propagates without dispersion.

Let $\mathcal{A}={\pi }^2\alpha \beta$ be the cross-sectional area. The appropriate time and cross-sectional average at fixed $x$ isWrite $F_{mn}(y,z)$ for the product of cosines and $Q={\mathcal{A}}^{-1}\int F_{mn}^{2}dydz$. Its transverse eigenvalue and wall conditions give $\int |{\nabla }_{\perp }{F}_{mn}{|}^2={\kappa }_{mn}^2\int F_{mn}^2$ by integration by parts. The kinetic energy density and compressive acoustic energy density are ${\rho }_0|\mathbf{u}{|}^2/2$ and $(p^{\prime }{)}^2/(2{\rho }_0{c}_0^2)$. Time averaging each squared real oscillation contributes a factor $1/2$, soThis holds for every dispersive mode, and also for the zero mode.

For localized smooth initial data, a dispersive mode contains integrals of the form $\int a(k)e^{it[kV-\omega (k)]}dk$ along $x=Vt$. When $0<V<c_0$, a stationary point exists at $k_{\ast }={\kappa }_{mn}V/\sqrt{c_0^2-V^2}$. It is nondegenerate because ${\omega }^{\prime \prime }(k)=c_0{\kappa }_{mn}^2/(k^2+{\kappa }_{mn}^2{)}^{3/2}>0$. The stationary phase method therefore gives an oscillatory amplitude $O(t^{-1/2})$ for each excited dispersive mode, and for their sum under sufficient mode summability.

At $V=c_0$, no dispersive mode has a finite stationary wavenumber; with smooth localized data these contributions decay, while the nondispersive right-moving pulse retains its original amplitude. Thus a general disturbance with nonzero zero-mode component has an undiminished $O(1)$ pulse on the sound-speed ray. If that component is absent, this surviving contribution is absent too. For $V>c_0$, finite propagation speed puts a compactly supported initial disturbance outside this ray for all sufficiently large $t$, so the field is eventually zero. Rapidly localized, rather than strictly compact, data give a correspondingly negligible tail instead.

For small $x$, use $u=\cosh t$ to write the defining oscillatory integral asThe first integral is $-\mathrm{Ci}(x)$, where the cosine integral satisfies $\mathrm{Ci}(x)={\gamma }_E+\log x+O(x^2)$. Its logarithm follows by integrating the $1/x$ singularity of its derivative; its conventional constant is the Euler--Mascheroni constant. The primary normalization is given by dlmf.nist.gov/6.6.E6 .

The second integrand has an integrable inverse-square-root endpoint singularity and an $O(u^{-3})$ tail. The dominated convergence theorem therefore gives its limitSplitting at $u=1/x$ and using $|1-\cos (xu)|\le \min (x^2{u}^2/2,2)$ also bounds its error by $O(x^2|\log x|)$. ThusIn particular, the leading term is $(2/\pi )\log x$.

For large $x$, the phase $\cosh t$ has its only stationary point at the endpoint $t=0$. Its expansion is $1+t^2/2+O(t^4)$, so the contributing width is $t=O(x^{-1/2})$. The endpoint stationary phase method and the Fresnel integral yieldA smooth cutoff near the endpoint justifies this local replacement; the remaining nonstationary tail can be integrated by parts, with no stationary contribution at infinity. Taking the real part gives the Bessel function of the second kind asymptoticThe additive error formulation remains meaningful near zeros of the leading oscillation. Both limits were derived from the given integral, rather than identifying a special function and merely quoting its asymptotics.

Initially suppose $0<\alpha <\pi /2$, the usual oscillatory range. The oscillatory phase is $n\Phi (\vartheta )$ with $\Phi =\sec \alpha \sin \vartheta -\vartheta$. Its stationary point is $\vartheta =\alpha$, since $\sec \alpha \cos \vartheta =1$. At that pointThe stationary phase method therefore gives the Debye Bessel asymptoticThe error is additive for fixed positive $\alpha$ bounded away from $\pi /2$; this formula is not uniform as $\alpha \to 0$, when the stationary point joins the endpoint and its curvature vanishes.

The printed condition $\alpha >0$ alone includes other trigonometric branches. All defined fixed real cases can be covered as follows. If $0<|\cos \alpha |<1$, put $\beta =\arccos |\cos \alpha |\in (0,\pi /2)$. Use the displayed formula with $\beta$ instead of $\alpha$, and multiply it by $1$ if $\sec \alpha >0$, or by $(-1{)}^n$ if $\sec \alpha <0$. This follows from the integer-order parity $J_n(-x)=(-1{)}^n{J}_n(x)$, which is also obtained from the defining integral by $\vartheta \mapsto \pi -\vartheta$. If $|\cos \alpha |=1$, the argument is $\pm n$ and the turning-point answer below applies, with the same parity factor. If $\cos \alpha =0$, the stated argument is undefined.

Apply the stationary phase method to the real part of the exponential integral for the Bessel function of the first kind. With $n$ fixed, use oscillatory phase $\sin \vartheta$ and oscillatory integral amplitude $e^{-in\vartheta }$. The unique stationary point is ${\vartheta }_0=\pi /2$, with second derivative $-1$. Its contribution isThe leading endpoint terms of the exponential integral are imaginary and do not contribute to its real part. Hence the large-argument asymptotic expansion of the Bessel function of the first kind isThis is an additive asymptotic formula: a relative ratio is inappropriate at zeros of the leading cosine. The envelope decreases like $x^{-1/2}$ while the oscillation frequency approaches one.

The linear elastic wave equation is $\rho {\mathbf{u}}_{tt}=(\lambda +\mu )\nabla (\nabla \cdot \mathbf{u})+\mu \Delta \mathbf{u}$. For a shear-horizontal wave, take $\mathbf{u}=(0,v(x,z,t),0)$ with no $y$ dependence. Its divergence vanishes, andRigid fixed boundaries require $v=0$ at $z=0,h$. The modes are $v=A\sin (q_{n}z)\cos (kx-{\omega }_{n}t)$ with $q_n=n\pi /h$, $n=1,2,\dots$, andThere is no $n=0$ displacement mode with these clamped boundary conditions. For $k>0$, the phase velocity and group velocity areThe phase velocity decreases from infinity to $c_s$, while the group velocity increases from zero to $c_s$; their product is $c_s^2$.

For a mode of frequency $\omega$, the appropriate time and cross-sectional average is $⟨A⟩=\omega /(2\pi h){\int }_0^{2\pi /\omega }{\int }_0^{h}Adzdt$, per unit transverse width. Its average kinetic energy density is $\rho A^2{\omega }^2/8$. The only nonzero strains are $e_{xy}=e_{yx}=v_x/2$ and $e_{yz}=e_{zy}=v_z/2$, while $e_{kk}=0$. Thus the elastic energy density is $\mu (v_x^2+v_z^2)/2$, whose average is $\mu A^2(k^2+q_n^2)/8$. The dispersion relation proves equality of average kinetic and elastic energies.

For the ray asymptotics of a clamped elastic waveguide mode, expand the localized initial displacement in transverse sine modes and Fourier transform in $x$. For a displacement released from rest, each mode evolves with $\cos ({\omega }_n(k)t)$. At $x=Vt$ the oscillatory phases are $t(kV\pm {\omega }_n(k))$. If $0<|V|<c_s$, one phase has a stationary point withThe stationary phase method gives a generic oscillatory amplitude proportional to $t^{-1/2}$ for each contributing mode. If the initial Fourier coefficient vanishes at the stationary point, that mode can decay faster; smooth localized data make the modal sum well behaved. If $|V|>c_s$, there is no stationary point. Repeated integration by parts gives rapid decay for smooth rapidly decaying data; for initially compactly supported displacement, finite propagation speed makes the displacement on that ray exactly zero after a sufficiently long time. These are the requested subsonic and super-shear-speed regimes.

Consider the permanent stationary topographic internal gravity wave component of a localized source, with constant buoyancy frequency in each layer and positive $U_1,U_2$. A sufficiently broad obstacle spectrum is assumed. The dispersion relation on the upward branch in a uniform layer isFor a stationary wave, $K=(k^2+m^2{)}^{1/2}=N/U$. Differentiating before imposing stationarity gives the laboratory group velocityThe vertical component is at most $U/2$, attained at $\theta =\pi /4$. The first arrival of a stationary internal wavefront at a layer boundary is thusIf the obstacle spectrum lacks that wavenumber, the first arrival of its actual permanent wave packets is later. This statement concerns the stationary part; arbitrary long-wave startup transients are not bounded by this particular envelope.

In the lower layer an elapsed propagation time $t$ gives $x=U_{1}t{\cos }^2{\theta }_1$, $z=U_{1}t\sin {\theta }_1\cos {\theta }_1$. Elimination gives the causal envelope of stationary internal wavesLater-emitted packets fill the interior. At $t=T$ this semicircle just touches $H$ at $x=H$. At $2T$ its unobstructed continuation would reach $2H$, but rays that hit the interface must instead be refracted or reflected.

For a lower-layer ray labelled by ${\theta }_1$, its interface arrival time and point areThe refraction of a stationary internal-wave causal envelope conserves horizontal wavenumber, so $\cos {\theta }_2=(U_2{N}_1)/(U_1{N}_2)\cos {\theta }_1$. If ${\tau }_1\le t$ and $0<\cos {\theta }_2<1$, its transmitted front is parametrized byThe reflected front instead has $x=U_{1}t{\cos }^2{\theta }_1$ and $z=2H-U_{1}t\sin {\theta }_1\cos {\theta }_1$, until it reaches the ground. At $2T$ that first return just touches the ground. If the $\pi /4$ incident ray is evanescent upstairs, upper-layer propagating arrival is delayed until an allowed ray reaches $H$; this does not change the first lower-layer arrival time $T$.

The stationary phase of topographic internal waves also gives the phase sketch. In a uniform lower layer the Fourier transform integrand has phase $kx+m(k)z$. The stationary phase method requires $x=kz/m$, giving phase $\Phi =(N_1/U_1)(x^2+z^2{)}^{1/2}$. Thus the dominant permanent phase lines are downstream circular arcs centred on the obstacle. Reflected phase arcs have the image source $(0,2H)$. Upstairs the ray arriving at $(x_H,H)$ has phasewhich is continuous at the interface and determines refracted constant-phase curves. Individual monochromatic components have straight phase lines of slope $-k/m_j$ instead; the arcs describe their localized-source superposition.

The schematic uses $H=U_1=N=1$ and $U_2=U_1/2$. It shows the direct front below $H$, the transmitted front above it and the reflected front below it; dotted arcs are permanent phase curves, not material trajectories.

The horizontal contribution is $B(x)=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }{e}^{-ix\sin z}{e}^{i\nu z}dz$. The two nondegenerate stationary points of the phase $-\sin z$ are $\pi /2$ and $-\pi /2$, with second derivatives $1$ and $-1$. The stationary phase method gives their contributionswith remainder $O(x^{-3/2})$ for smooth fixed amplitudes localized near the stationary points. On the remaining portions, use integration by parts. The endpoint term isCutoff endpoints cancel between pieces; a second integration gives the other nonstationary contributions $O(x^{-2})$. Thus this endpoint term cancels the leading vertical term. AltogetherThis stationary phase endpoint cancellation in a half-strip contour is necessary: omitting it would leave an erroneous $x^{-1}$ remainder for noninteger $\nu$.

For a Fourier mode $e^{i(kx-\omega t)}$, the wave equation givesTaking the positive-frequency branch,For $k>0$, the phase velocity and group velocity areThe frequency rises from zero and approaches $1$; $c_p$ reaches its maximum at $k=1$, whileshows that the front velocity isThus the wave at the front hasMoreover $c_g/c_p=2/(1+k^2)$, so the packet outruns its crests for $0<k<1$, the speeds agree at $k=1$, and the crests outrun the packet for $k>1$.

The initial conditions give the exact superpositionThe condition $A^{\ast }(-k)=A(k)$ makes this solution real. For $V=x/t$ with $0<V<V_m$, the equationhas two positive roots $0<{\alpha }_1<1/\sqrt{3}<{\alpha }_2$. The second exponential has the corresponding stationary points $-{\alpha }_1,-{\alpha }_2$. Sincethe stationary phase method and pairing of complex-conjugate contributions giveFor $-V_m<V<0$, the same picture is reflected and the stationary points have negative wavenumber. For $|V|>V_m$ there are no real stationary points, so a smooth localized spectrum gives a much smaller nonstationary tail outside the wave front.

Each Fourier mode has zero initial velocity, soAlong $x=Vt$ this isPutThe two phases have stationary points at $-k_0$ and $k_0$, respectively. At those points,

Because the initial field is real, $a(-k)=a(k{)}^{\ast }$. Applying the stationary phase method to the two conjugate contributions gives the stationary-phase asymptotic of a Klein-Gordon wave along a subluminal ray:Equivalently, if $a(k_0)=|a(k_0)|e^{i\delta }$,

The only nondegenerate stationary point of a phase of the phase $t^2$ is the endpoint zero. Its usual leading stationary phase method term is proportional to the amplitude there, but ${\tan }^2(0)=0$. Thus the nominal $x^{-1/2}$ stationary contribution vanishes.

Expand one order further in the local amplitude: ${\tan }^{2}t=t^2+O(t^4)$. The rescaling $t=u/\sqrt{x}$ then gives a first nonzero stationary contribution of order $x^{-3/2}$, with coefficient $-\sqrt{2\pi }/8$ in the real integral, as in part (i). Meanwhile the upper endpoint is nonstationary and has nonzero amplitude; its integration by parts contribution is $2\sin ({\pi }^{2}x/16)/(\pi x)$, of order $x^{-1}$.

Consequently the nonstationary endpoint dominates the first nonzero saddle contribution. A stationary phase method formula retaining only the usual leading saddle term would miss the answer. The full stationary phase method, with the vanishing-amplitude correction and the endpoint term, agrees with method of steepest descent. This is quadratic stationary endpoint with quadratically vanishing amplitude.

Write the cosine in i as two complex exponentials. On a ray $x=vt$, the oscillatory phases are $kv\pm \omega (k)$. The stationary phase method requires $v=\mp {\omega }^{\prime }(k)$. By ii, the largest absolute horizontal group velocity for the occupied $n=1$ mode isInside this region there are stationary wavenumbers supporting appreciable wave packet contributions. Outside it no stationary wavenumber exists; oscillatory cancellation suppresses the large-time response. This explains "mainly confined": the initial Lorentzian profile already has nonzero algebraic tails, so the displayed region is not the exact support of the solution. Near its boundary the stationary points meet $k=0$ and the usual nondegenerate stationary-phase formula needs a transition approximation.

The stationary endpoint contributes $c{e}^{3i\pi /4}\sqrt{\pi }/(4x^{3/2})$ by method of steepest descent along $t=e^{i\pi /4}r$. Its ordinary $x^{-1/2}$ term vanishes because $a(0)=0$. A nonstationary upper endpoint with $a(b)\ne 0$ contributes $a(b)e^{ix{b}^2}/(2ibx)$ and therefore dominates the stationary contribution. The stationary phase method must retain both this boundary term and the first nonzero local-amplitude term.