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Fluid-loaded plate. Use complex amplitudes with time dependence $e^{-i\omega t}$ and take the velocity potential in the fluid above the elastic plate to be $\Phi$. Let ${\rho }_0$ and $c_0$ be the undisturbed mass density and speed of sound. The linearized Euler equations give $p=i{\rho }_0\omega \Phi$; the kinematic boundary condition is ${\Phi }_y(x,0)=-i\omega \eta (x)$. We model loading from one fluid half-space, with no fluid loading on the other face. Loading by identical fluids on both faces instead doubles the fluid term below.

Define the Fourier transform and its inverse byThe Sommerfeld radiation condition selects $\hat{\Phi }(k,y)=\hat{\Phi }(k,0)e^{-\gamma y}$, with ${\gamma }^2=k^2-k_0^2$ and $k_0=\omega /c_0$. Here and in the diffraction calculation below, the outgoing square root is specified by first taking $\mathrm{Im}\omega >0$. Draw a vertical cut upwards from $+k_0$ and another downwards from $-k_0$, and putThus ${\gamma }_u$ is analytic in the upper half-plane and ${\gamma }_l$ in the lower half-plane. In the real-frequency limit,The Fourier transform contour $C$ follows the real axis with the limiting-absorption pole prescriptions. In particular, an outgoing propagating component has positive vertical wavenumber.

The kinematic boundary condition gives $\hat{\Phi }(0)=i\omega \hat{\eta }/\gamma$, so $\hat{p}(0)=-{\rho }_0{\omega }^2\hat{\eta }/\gamma$. Eliminating the fluid pressure from the bending balance gives the line-forced fluid-loaded bending plate kernelFor harmonic line force amplitude $F_0$ at the origin, the plate displacement and fluid velocity potential are thereforeAn arbitrary localized loading replaces $F_0$ by its Fourier transform. This representation both calculates the response and identifies its different wave components through poles and branch points.

For an observer $x=r\cos \theta$, $y=r\sin \theta$, with $0<\theta <\pi$, the method of steepest descent uses the saddle point $k_s=k_0\cos \theta$. At fixed angle away from grazing, the acoustic far field isThe $r^{-1/2}$ spreading is that of cylindrical acoustic waves. The phase velocity parallel to the plate is $\omega /|k|$: components with $|k|<k_0$ are supersonic in that sense and radiate, whereas those with $|k|>k_0$ are subsonic and exponentially decay normally to the plate. The latter have reactive fluid loading and zero mean normal acoustic energy flux.

There is also a real guided root $\kappa >k_0$. Indeed, $K(k_0+)=-\infty$, $K(+\infty )=+\infty$, andThe residue theorem then gives an outgoing displacement contribution $i{F}_0{e}^{i\kappa |x|}/K^{\prime }(\kappa )$. It transports energy along the plate while its fluid field decays as $e^{-\gamma (\kappa )y}$. Other zeros on continued sheets can describe leaky or resonant waves; branch point contributions describe near and grazing fields. A fixed-angle acoustic far field expansion is not uniform near the plate, where these components can matter.

For a propagating spectral component, write $\ell =(k_0^2-k^2{)}^{1/2}$. Its upward normal velocity is $i\ell \hat{\Phi }$ and its pressure is $i{\rho }_0\omega \hat{\Phi }$. Applying Parseval's identity to the time-averaged acoustic energy flux gives the upward power per unit length of the forcing line:Equivalently, the integrand is $|\hat{\eta }{|}^2/\ell$ with prefactor ${\rho }_0{\omega }^3/(4\pi )$. Integrating the radial acoustic intensity of the far field over a semicircle gives the same expression. The input power $\omega \mathrm{Im}(F_0^{\ast }\eta (0))/2$ also feeds guided waves along the plate, so it must not be identified with $P_y$ alone.

Rigid half-plane diffraction. Suppress the time factor and assume $0<{\theta }_0<\pi$. Write $a=k_0\cos {\theta }_0$, $b=k_0\sin {\theta }_0$, and denote the incident spatial field by $I=e^{-iax-iby}$. The Helmholtz equation, Sommerfeld radiation condition, and square-root branches are as above. Across the open positive axis the scattered field and its normal derivative are continuous. Across the rigid negative axis its normal derivative is also continuous, while its potential can jump. Let $J(x)$ be this jump, supported on $x<0$. The transformed scattered field isThe jump transform is analytic in the upper half-plane. The absence of an even scattered component follows from derivative continuity and the outgoing condition. Its normal derivative on either face is $-\gamma {\hat{J}}_u/2$.

On the negative axis this derivative must cancel $-ib{e}^{-iax}$. Since the Half-range Fourier transform of $H(-x)e^{-iax}$ is $i/(k+a)$, the Wiener-Hopf equation has the formwhere $G_l$ is the transform of an unknown function supported on $x>0$ and analytic in the lower half-plane. The incident pole $k=-a$ lies below the Fourier contour. Divide by ${\gamma }_l$ and perform pole subtraction in a Wiener-Hopf equation:The first term is upper analytic and the second lower analytic, since the latter's apparent pole cancels. Equality of the two continued sides gives an entire function. The finite-energy edge condition, with $J(x)=O(\sqrt{-x})$, and the decay of the transforms at infinity force that function to vanish by Liouville's theorem. ConsequentlyThis is the Wiener-Hopf solution of rigid half-plane diffraction with the present transform and time conventions.

Deforming to a steepest descent contour gives a residue from the incident pole precisely when $x+|y|\cot {\theta }_0<0$. At that pole $\gamma (-a)=-ib$; its residue producesThus the total geometrical optics field is $I$ plus a reflected wave in the upper reflection region; in the lower shadow region the scattered residue cancels $I$, and outside that shadow the direct incident wave remains. This gives both the amplitude and the spatial support of the geometrical optics fields rather than merely naming a reflected wave.

For $x=r\cos \theta$, $y=r\sin \theta$, $-\pi <\theta <\pi$, the saddle contribution uses ${\gamma }_l(-a)=-i\sqrt{2k_0}\cos ({\theta }_0/2)$ and ${\gamma }_u(k_s)=\sqrt{2k_0}\cos (\theta /2)$. The diffracted wave isThe total field is $I+{\varphi }_{\mathrm{GO}}+{\varphi }_d$ to this order. This far-field expression applies away from the shadow/reflection boundaries $|\theta |=\pi -{\theta }_0$. There the pole and saddle coalesce, and a uniform Fresnel integral approximation replaces the divergent separate terms by a smooth transition. The exact contour integral remains well defined.

First take $z>0$. Set $s=t^3-1$ in the tail integral. This puts it in the endpoint form needed for Watson's lemma:The binomial coefficients are $(-1{)}^r(2/3{)}_r/r!$, where $(a{)}_r=\Gamma (a+r)/\Gamma (a)$ is the rising factorial. Integrating each endpoint term givesAlso, scaling the complete positive-ray integral gives ${\int }_0^{\infty }{e}^{-z^3{t}^3}dt=\Gamma (1/3)/(3z)$. HenceThe endpoint term is exponentially small on this positive ray but must be retained for continuation to other sectors.

For complex $z$, use $s=zt$ so $G(z)=z^{-1}{\int }_0^z{e}^{-s^3}ds$. Put $C=\Gamma (1/3)/3$ and $\omega =e^{2\pi i/3}$. The complete steepest descent contours from the cubic saddle are the rays $\arg s=2\pi j/3$, whose integrals are $C{\omega }^j$. Deform the finite contour onto one of these rays followed by the endpoint contour. The cubic exponential integral Stokes sectors giveFor $0\le \arg z\le 2\pi$, take $j=0$ below $\pi /3$, $j=1$ between $\pi /3$ and $\pi$, $j=2$ between $\pi$ and $5\pi /3$, and $j=3$ above $5\pi /3$. The corresponding algebraic coefficients are $C,C\omega ,C{\omega }^2,C$. A negative-real $z$ provides a useful sign check: $G(z)=G(|z|e^{i\pi })$ is real and positive, so its two lateral algebraic coefficients must be complex conjugates.

The subdominant algebraic term switches across the Stokes linesThere $z^3$ is negative real and the endpoint exponential is dominant. The intervening anti-Stokes lines are $\arg z=\pi /6+j\pi /3$, where $\mathrm{Re}{z}^3=0$ and the exponential changes from growth to decay. The naming convention here distinguishes switching from equal exponential magnitude.

To resolve the jump near ${\theta }_s=\pi /3$, optimally truncate the endpoint series at $N$ with $N+2/3=R+O(1)$, where $R=|z{|}^3$. Let $P_N$ denote that truncated endpoint contribution. The Borel remainder for a cubic exponential integral derived in solution, together with the Gaussian pole transition for a large gamma parameter in solution, givesThis estimate is on the transition scale $\theta -\pi /3=O(R^{-1/2})$ after optimal truncation. The error-function smoothing of a Stokes multiplier has width $O(|z{|}^{-3/2})$. Before the line $S\to 0$ and the coefficient is $C$; after it $S\to 1$ and the coefficient is $C\omega$, agreeing with the sector expansions. On the line the leading coefficient is their average. The original $G$ is an entire function; only its asymptotic decomposition has these sector-dependent coefficients.

Take $0<{\theta }_0<\pi$, put $a=k_0\cos {\theta }_0$ and $b=k_0\sin {\theta }_0>0$, and suppress the common factor $e^{i\omega t}$. The upper-half-plane incident acoustic plane wave is $e^{iax+iby}$. A rigid zero-thickness screen imposes the same normal-derivative cancellation on both faces; the scattered velocity potential is consequently odd in $y$. Across the open part $x>0$ it is continuous, so its trace there is zero. These complementary half-line data are what make the Wiener-Hopf method applicable.

Use the Fourier transform $\hat{\varphi }(\kappa ,y)={\int }_{\mathbb{R}}\varphi (x,y)e^{i\kappa x}dx$, with inverse exponential $e^{-i\kappa x}$. The outgoing Helmholtz equation solution is $\hat{\varphi }=F^{-}(\kappa )e^{-\gamma y}$ for $y>0$, where $\gamma =\sqrt{{\kappa }^2-k_0^2}$ has $\gamma =i\sqrt{k_0^2-{\kappa }^2}$ on the propagating interval and positive real part on evanescent components. The limiting absorption principle supplies the contour prescription. The transform $F^{-}$ of the trace supported on $x<0$ is analytic below the contour. Let $G^{+}$ be the analytic upper transform of the unknown derivative on $x>0$. The prescribed derivative on $x<0$ has transformSince the upper normal derivative is $-\gamma F^{-}$, the Wiener-Hopf equation isFor the Wiener-Hopf factorization, choose ${\gamma }_{+}=(\kappa -k_0{)}^{1/2}$ analytic above and ${\gamma }_{-}=(\kappa +k_0{)}^{1/2}$ analytic below. Their branch cuts go into the opposite half-planes, with ${\gamma }_{+}(-a)=i\sqrt{k_0+a}$. Dividing by ${\gamma }_{+}$ and performing pole subtraction in a Wiener-Hopf equation givesThe forcing pole is allocated to the lower analytic part; the second quotient has a removable pole and is upper analytic. Subtracting these parts leaves a common entire function. The finite-energy edge condition excludes an inverse-square-root singularity in the potential: its odd scattered trace is $O(\sqrt{|x|})$, so $F^{-}=O(|\kappa {|}^{-3/2})$. The normal derivative on the open side can be $O(x^{-1/2})$, giving $G^{+}=O(|\kappa {|}^{-1/2})$. Both sides of the entire remainder tend to zero. The Liouville theorem sets that remainder to zero, giving the explicit Wiener-Hopf solution of rigid half-plane diffraction:The scattered field on either side of the screen is thereforeIts forcing-pole residue at $\kappa =-a$ is $-i$. Deforming to the outgoing steepest descent contour crosses this pole precisely when $|\theta |>\pi -{\theta }_0$, where $x=r\cos \theta$, $y=r\sin \theta$ and $-\pi <\theta <\pi$. The residue contributes $\mathrm{sgn}(y)e^{iax-ib|y|}$. Thus, away from transition boundaries, the scattered geometrical optics field isAdding the incident acoustic plane wave gives incident plus unit specular reflection in the upper reflected sector, just the incident wave in the remaining illuminated sectors, and zero in the lower shadow sector. The geometrical shadow boundaries also follow directly by tracing a straight ray to $y=0$: the reflected ray hits the screen at $x+y\cot {\theta }_0<0$, and the transmitted incident ray crosses the open part at $x-y\cot {\theta }_0>0$.

For the diffracted contribution, the outgoing phase $-i\kappa r\cos \theta -\gamma r|\sin \theta |$ has saddle ${\kappa }_s=k_0\cos \theta$, value $-i{k}_{0}r$, and second derivative $ir/(k_0{\sin }^2\theta )$. Stationary phase therefore supplies $\sqrt{2\pi k_0/r}|\sin \theta |e^{-i{k}_{0}r+i\pi /4}$. Substituting $F^{-}({\kappa }_s)$, and using $\mathrm{sgn}(y)|\sin \theta |/\sqrt{1+\cos \theta }=\sqrt{2}\sin (\theta /2)$, givesThe full asymptotic scattered field is ${\varphi }_{\mathrm{GO}}+{\varphi }_d$. The angular edge-wave expression is not uniform on the geometrical boundaries. There is also a sign inconsistency in the supplied saddle formula: with its printed inverse exponential $e^{+i\kappa x}$, the outgoing saddle is $-k_0\cos \theta$, not $+k_0\cos \theta$. The consistent inverse convention used above gives the requested outgoing field without that ambiguity.

Take the principal complex logarithm and deform the interval upwards, where $e^{ixt}$ decays for $x>0$. The small indentation at zero contributes nothing because $r\log r\to 0$; the distant horizontal segment vanishes exponentially. The two vertical edges therefore give the exact representationThis is an endpoint steepest descent contour, with distinct contributions from the singular endpoint and the smooth endpoint. Setting $u=xs$ in the first integral and using the given logarithmic moment givesFor the second edge, the Taylor series at $s=0$ is $\log (1+is)={\sum }_{n\ge 1}(-1{)}^{n+1}(is{)}^n/n$. Watson's lemma integrates it term by term as an asymptotic expansion, using ${\int }_0^{\infty }{s}^n{e}^{-xs}ds=n!/x^{n+1}$. ConsequentlyHere $\gamma$ is the Euler--Mascheroni constant. The first smooth-endpoint terms areFor every fixed $N\ge 1$, retaining $k=0,\dots ,N-1$ leaves $O(x^{-N-2})$. The factorial series is an asymptotic series, not a convergent sum. Treating the logarithmic endpoint by an ordinary smooth-endpoint integration formula would lose the leading logarithm; this is the logarithmic endpoint Fourier asymptotics mechanism.

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.

Use the entire function $e^{ik{z}^3}$. The steepest descent contour starting at $1$ iswith the cube-root branch starting at $1$. Along it $ik{z}^3=ik-ku$, so the real part decreases and the imaginary part stays constant. It approaches the ray $\arg z=\pi /6$, on which $z=e^{i\pi /6}{u}^{1/3}$ and $ik{z}^3=-ku$.

Close a contour using the interval $[0,1]$, these two outgoing paths truncated at parameter $R$, and a short connecting arc. The two endpoints differ by $O(R^{-2/3})$, and the integrand on the connector decays exponentially in $R$. Cauchy integral theorem therefore gives ${\int }_0^1{e}^{ik{t}^3}dt=I_1-I_2$.

The specified substitutions giveandExpand $(1+iu{)}^{-2/3}=1-2iu/3+O(u^2)$ near zero, and apply Watson lemma, or split the integral at a fixed small $u$ and integrate the Taylor remainder. This gives the first two asymptotic termsKeeping the next term gives $-2e^{ik}/(9k^2)+O(k^{-3})$. The fractional power comes from the degenerate stationary endpoint at zero; the oscillatory integer-power terms come from the ordinary endpoint at one.