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Use the Fourier transform pair $\hat{q}(k)=\int q(x)e^{ikx}dx$, $q(x)=(2\pi {)}^{-1}\int \hat{q}(k)e^{-ikx}dk$. The point force transforms to $F$. With the dispersion relation $D$ defined above, the sheet equation becomesThe point-force radiation from a fluid-loaded sheet is therefore represented exactly byThe causal contour and outgoing acoustic square-root branch are fixed first with $\mathrm{Im}\omega <0$, then continued to the desired real frequency. This prescription fixes how poles and the branch points are passed.

For the acoustic far field $x=r\cos \theta$, $y=r\sin \theta$, take $0<\theta <\pi$ bounded away from grazing and $k_{0}r\gg 1$. The method of steepest descent saddle point is $k_s=k_0\cos \theta$, with ${\gamma }_s=i{k}_0\sin \theta$. The supplied saddle point rule, including its $\sin \theta$ factor, gives, provided the contour deformation crosses no poles,A convenient simplification, free of division by $T$, isEquivalently, when $T\ne 0$,The expression printed in the PDF is missing sound-speed factors for general dimensional $c_0$. It agrees with this result if $c_0=1$ in fully normalized units; when $c_0$ is retained as an arbitrary sound speed, the numerator needs $c_0^{-3/2}$ and the structural term needs $T/c_0^2$ in the last form. These factors arise respectively from cylindrical spreading, the pressure-density relation, and $k_s=\omega \cos \theta /c_0$.

A direct countercheck is the transparent-sheet limit $m=T=0$. The sheet jump condition then gives ${\hat{p}}_{+}(k,0)=F/2$, so the saddle point rule requiresThe printed expression, interpreted continuously after multiplying out its structural factor, instead gives $F(2c_0{)}^{-1}\sqrt{\omega /(2\pi r)}\sin \theta$ times the same phase. It differs by a factor $c_0^{3/2}$; for example it is eight times too large when $c_0=4$. This limit also verifies the normalization of the corrected density field independently of the sheet's elastic-sheet tension.

To decide about poles, track the roots of $\Delta (k,\omega )$ on the chosen square-root sheet and deform the original causal contour to the steepest descent contour. A root contributes a residue exactly when it lies in the region swept out by that deformation; its sign is fixed by the contour orientation. Branch cuts must be retained throughout this comparison. Which roots are crossed can depend on observation angle, producing a change of the modal contribution when a pole meets the deformation boundary. A saddle point approaching a pole or a grazing endpoint requires an approximation uniform in that limit, rather than the isolated saddle point formula above.

The crossed poles are the free fluid-sheet modes of the preceding solution. Real subsonic roots represent evanescent acoustic surface waves carrying energy along the sheet, with normal decay; complex continuations represent leaky or radiating modes. Their residues must be added to the saddle point sound when the causal contour selects them. The specification “no poles contribute” is therefore a substantive condition on the contour, not permission to ignore zeros of the dispersion relation.

Let $b=e^{i\theta }$ and use the method of steepest descents with phase $\Phi (z)=-z^2$. For $z=x+iy$, $\mathrm{Re}\Phi =y^2-x^2$ and $\mathrm{Im}\Phi =-2xy$. The simple saddle point is $z=0$: the real axis is its descending path and the imaginary axis its ascending path. The quadratic complex endpoint descent path from $b$ satisfies $\mathrm{Im}(z^2)=\sin 2\theta$ and can be parametrized bywith the square-root branch chosen continuously from $b$. Increasing $s$ reduces the real part of the exponent by $\lambda s$.

For $0\le \theta <\pi /2$, that endpoint path goes to positive real infinity. Deform the original contour into the real path from $-d$ through the saddle point towards $+\infty$, followed backwards along the endpoint descent path to $b$. The integrand is an entire function, so finite contour deformations are legitimate; connecting arcs in the far decaying sector vanish. If $T_b$ denotes the oriented integral from $b$ to that infinity, this givesThe last expression follows by setting $z^2=b^2+s$: $dz/ds=1/(2z)$, so the endpoint integral is $e^{-\lambda b^2}{\int }_0^{\infty }{e}^{-\lambda s}/[2z(s)]ds$. This derives its phase and orientation directly.

For $\pi /2<\theta \le \pi$, the descent path from $b$ goes to negative real infinity. Deform from $-d$ towards that infinity and then backwards along this endpoint path to $b$. There is no saddle point segment, anduses the same formula with the negative square-root branch. At $\theta =\pi /2$ the contour can run from $-d$ to zero and then up the imaginary axis. It gives a half-Gaussian contribution and the dominant imaginary endpoint term $i{e}^{\lambda }/(2\lambda )$. The change in the subdominant saddle point contribution is a Stokes phenomenon; it does not change the leading endpoint term on either side of that angle.

The Gaussian saddle and endpoint dominance by angle follows by comparing exponential sizes and gives all fixed-angle leading behaviours:At $\theta =\pi /4$, the endpoint term has magnitude $O({\lambda }^{-1})$, so the saddle point term of order ${\lambda }^{-1/2}$ still dominates. In the middle interval the endpoint term grows exponentially for $\pi /4<\theta <3\pi /4$, is oscillatory of order ${\lambda }^{-1}$ at $3\pi /4$, and decays exponentially for $3\pi /4<\theta <5\pi /6$. On the latter interval it nevertheless decays more slowly than the fixed-endpoint term. Equality of their exponential rates means $\cos 2\theta =1/2$ on the left half-plane, hence $\theta =5\pi /6$. At that angle both terms are required; their coefficient magnitudes are unequal, so their sum has no complete destructive cancellation.

These fixed-angle asymptotic expansions are not uniform if $\theta$ is allowed to vary with $\lambda$ near a dominance transition. The endpoint/saddle balance near $\pi /4$ involves $\lambda (\theta -\pi /4)=O(\log \lambda )$; the two-endpoint competition near $5\pi /6$ has an angular scale $O({\lambda }^{-1})$.

The following sketches show the contour orientations, the saddle point and both endpoints. The hyperbolic endpoint paths are followed backwards when returning from infinity to $b$; the dotted far-end connectors indicate the vanishing connecting arc.

First make explicit the usual method of steepest descent hypotheses: the saddle point is simple, $G^{\prime }(z_0)=0$ and $g_2=G^{\prime \prime }(z_0)\ne 0$; a contour deformation leads to the two descent arms with one clockwise half-indentation around the pole; and endpoints or other contour portions contribute smaller terms. The sign of the exponential matters: $|e^{-i\lambda G}|=e^{\lambda \mathrm{Im}G}$, so decay is toward valleys of $\mathrm{Im}G$.

Use a local analytic function $w(z)$ as coordinate. Nondegeneracy allows an analytic square root after factoring the double zero of $G-G(z_0)$, so the coordinate can be chosen withThe sign of $c$ is selected by the orientation of the contour integral, with $w$ running from negative to positive real values. Matching cubic coefficients gives $d/c=-c{G}^{\prime \prime \prime }(z_0)/(6g_2)$. The transformed amplitude therefore has the Laurent seriesThe Cauchy principal value contribution of the pole on the two real arms is zero by oddness. Its clockwise semicircular indentation contributes $-i\pi F(z_0)$, since $\int dw/w=-i\pi$ there. The regular term contributes the Gaussian integral $B\sqrt{2\pi /\lambda }$. Consequently the simple saddle coinciding with a simple pole has the additive expansionThe remainder written here assumes the other contour contributions are negligible to that order. Otherwise the local result has to be supplemented by them. The next local correction is generically $O({\lambda }^{-1/2})$ times the displayed exponential. If $B=0$, it can vanish; odd Taylor terms integrate to zero between the symmetric arms. When $F(z_0)\ne 0$, the leading term gives the printed asymptotic equivalence. When $F(z_0)=0$, asymptotic equivalence to zero is not meaningful: the additive formula is still valid and a regular saddle-point approximation may lead instead.

Nondegeneracy is necessary and is not explicitly printed. For a counterexample to the unrestricted reading, take $F=1$, $z_0=0$ and $G(z)=-i{z}^4$. There is one critical point, but it is degenerate. Choose the contour from $+i\infty$ to $+\infty$, with a clockwise quarter-circle avoiding zero. Both rays are decaying valleys for $e^{-\lambda z^4}$. Their straight-ray contributions cancel, and the indentation gives $J=-i\pi /2$, not $-i\pi$. Thus the requested coefficient requires the simple saddle point and contour deformation assumptions above; the mere existence of one critical point is insufficient.

The exponent $\Phi (t)=\frac{i}{2}(t-t^{-1})$ has saddles $t=\pm i$, with $\Phi (-i)=1$ and $\Phi (i)=-1$. Since the integrand is analytic off the origin, deform the contour to the unit circle and rotate $t=-i{e}^{i\theta }$. This gives the exact representationThe dominant saddle at $\theta =0$ is a steepest-descent direction for the method of steepest descent: $\cos \theta =1-{\theta }^2/2+{\theta }^4/24-\cdots$. Set $\theta =s/\sqrt{x}$ and expand the analytic amplitude against $e^{-s^2/2}$. Integrating its Gaussian moments gives the leading factor $e^x/\sqrt{2\pi x}$; the rest of the contour is exponentially smaller than this dominant contribution outside any fixed neighbourhood of the maximum.

To obtain every coefficient efficiently, the integral satisfies $x^2{I}_0^{\prime \prime }+x{I}_0^{\prime }-x^2{I}_0=0$, by differentiating and integrating the derivative of $\sin \theta e^{x\cos \theta }$. Substitute $I_0=e^x{x}^{-1/2}\sum a_n{x}^{-n}/\sqrt{2\pi }$ to get $a_0=1$ and $a_n=(2n-1{)}^2{a}_{n-1}/(8n)$. Therefore the full asymptotic expansion isHere $(-1)!!=1$. For each fixed truncation at $n=N$, the relative remainder is $O(x^{-N-1})$ as $x\to +\infty$. The infinite series is generally divergent, and this all-orders expansion does not purport to resolve exponentially small saddle contributions or their Stokes phenomenon.

Choose the square root with $-\pi /2<\arg z<3\pi /2$, so the branch cut is the negative imaginary axis. The change of variable $u=e^{-i\pi /4}\sqrt{z}$ maps this sheet onto $\mathrm{Re}u>0$ and gives $z=i{u}^2$. Writing the exponential function as $e^{k\varphi (z)}$, its phase becomes exactly quadratic:The saddle point is therefore $u=1$, or $z_s=i$. On the descending line $u=1+iv$, $v\in \mathbb{R}$, the phase is $-1-v^2$. Its image is the parabolaThe contour runs from left to right, corresponding to $v$ decreasing from $+\infty$ to $-\infty$. A contour deformation in the right half of the $u$-plane moves the original indented contour to this line. The connecting tails vanish in the descending sectors, and no branch point or pole is crossed. Since $dz=-2(1+iv)dv$, the Gaussian integral and the odd function $v{e}^{-k{v}^2}$ giveHere the method of steepest descent actually gives an exact answer for $k>0$, because the quadratic phase and linear transformed amplitude have no further even correction.

Put $t=u+iv$ and $\Phi (t)=\sinh t-t$. Sincea steepest-descent curve through the saddle point $t=0$ lies inThe saddle is cubic rather than simple:Thus the curves meet the origin in the three descent directions for which $t^3$ is negative real,One curve is the negative real axis $v=0$, $u<0$. The other two are conjugate curves $v=\pm v_{+}(u)$ with $u>0$ and $0<v_{+}(u)<\pi$. Near the origin, $v_{+}(u)\sim \sqrt{3}u$, while for $u\to +\infty$ the defining equation gives $\sin v_{+}\sim 2v_{+}{e}^{-u}$, so $v_{+}\to \pi$. Their three asymptotes are therefore

PutOn the specified branch, $\sqrt{z^2-1}$ is positive for $z>1$ and negative for $z<-1$. The saddle points of the phase are therefore $z=\pm a$, withandApplying the method of steepest descent at the two simple saddles givesThis formula is valid while the pole and positive saddle remain separated by much more than their $O({\lambda }^{-1/2})$ saddle width.

The pole crosses the positive saddle whenBecause the original contour passes above the pole, deformation onto the steepest-descent contour contributeswhen $z_0>a$, or $0<\theta <{\theta }_{\ast }$. It contributes no residue when $z_0<a$, or ${\theta }_{\ast }<\theta <1$. Hence, away from the transition,The residue is exponentially oscillatory and $O(1)$, whereas each ordinary saddle contribution is $O({\lambda }^{-1/2})$.

At $\theta ={\theta }_{\ast }$ the pole and saddle coalesce, so the displayed saddle formula is singular and must not be used. Passing above the coincident point gives one half of the switched residue at leading order:If $|\theta -{\theta }_{\ast }|=O({\lambda }^{-1/2})$, a uniform saddle-point approximation with a nearby pole replaces the discontinuous switch by a complementary-error-function multiplier.

As $\theta \to 1^{-}$, $s\to 0$, $a\to \infty$, and the pole lies in the no-residue regime. Each saddle coefficient is $O(s^{1/2}{\lambda }^{-1/2})$, so the leading approximation tends to

For $F(t)=\mathrm{sech}\alpha \sinh t-t$, the relevant saddle point is $t=\alpha$, where $F(\alpha )=\tanh \alpha -\alpha$ and $F^{\prime \prime }(\alpha )=\tanh \alpha$. The method of steepest descent gives

Write the phase and amplitude asThe saddle point relevant to the contour is $z=0$, since ${\varphi }^{\prime }(z)=\sinh z$ andThe local steepest-descent direction is therefore vertical: with $z=iy$,

The given contour approaches the lines $\mathrm{Im}z=-\pi$ at its left end and $\mathrm{Im}z=\pi$ at its right end. The poles of $f$ are at $z=\pm 4i$, outside the closed strip $|\mathrm{Im}z|\le \pi$. The Cauchy integral theorem therefore permits deformation to the contour consisting of the lower horizontal ray from $-\infty -i\pi$ to $-i\pi$, the vertical segment from $-i\pi$ to $i\pi$, and the upper horizontal ray from $i\pi$ to $\infty +i\pi$. On either horizontal ray,so those two integrals are exponentially small, of order $e^{-x}{x}^{-1/2}$.

The vertical segment, oriented upward, contributesIts real phase has a unique maximum at $y=0$. Sincethe simple-saddle contribution in steepest descent, equivalently the local Gaussian integral, givesThe factor $i$ comes from the upward tangent $dz=idy$ of the deformed contour.

The exponential phase isIts saddle points satisfyFor positive real $\lambda$, the original positive-real contour passes through $s=\lambda$; the singularity at $s=0$ separates it from the other saddle. At the contributing saddle,Set $s=\lambda +{\lambda }^{1/2}t$. ThenThe method of steepest descent therefore gives

The endpoint $t=0$ is treated by rotating onto its method of steepest descent ray. It contributesAt $t=1$, put $s=1-t$ and expandTermwise endpoint integration, using ${\int }_0^{\infty }{s}^n{e}^{-ixs}ds=n!(ix{)}^{-n-1}$ on the rotated ray, giveswhereThe two displayed contributions come from the two endpoints; the remaining deformed contour is exponentially or algebraically smaller after any fixed truncation.

At the two saddles,andApplying the method of steepest descent with the oriented angles from part (ii), and including the prefactor $1/(2^{n+1}\pi i)$, gives two conjugate contributions. Iftheir sum isTherefore the Debye asymptotic for Legendre polynomials isThus

Write the Schläfli contour integral for Legendre polynomials asThe saddle points satisfyFor $\mu =\cos \theta$, they are $z_{\pm }=e^{\pm i\theta }$. Deforming the contour through both conjugate saddles and using the local Gaussian contributions from the method of steepest descent gives conjugate exponentials. Their sum is the Debye asymptotic for Legendre polynomialsBoth saddles are required for the real oscillatory answer.

A harmonic line force localized at one point of a two-dimensional sheet gives $\hat{\eta }=F/D$ and $\Delta =\gamma D$ in the Fourier transform. The outgoing upper pressure is the displayed amplitude. Dividing by $c_0^2$ converts it to acoustic density perturbation. The method of steepest descent gives cylindrical sound from the saddle point, while crossed zeros of the dispersion relation give guided or leaky modal residues. Retaining $k_0=\omega /c_0$ is essential for dimensionally consistent spreading and prefactors.