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For the Airy equation with opposite dispersive sign, $q_t-q_{xxx}=0$ on $x>0$, let $Q$ be the initial Half-range Fourier transform and $F_j(k,t)={\int }_0^t{e}^{i{k}^{3}s}{\partial }_x^{j}q(0,s)ds$. Three integrations by parts give the displayed global relation. The spatial transform is holomorphic for $\mathrm{Im}k<0$, whereas the finite-time spectral boundary transforms are entire functions. The symmetry $k\mapsto \alpha k$, ${\alpha }^3=1$, preserves their time exponent.

// Target: analysis.bigb

A pinned semi-infinite membrane scatters an incoming guided mode into reflected guided modes and outgoing sound. For unit incident displacement the scattered endpoint value is $-1$. Its Half-range Fourier transform therefore has nonzero endpoint terms even though the total displacement vanishes. The Wiener-Hopf equation kernel is $2/(m{\omega }^2-T{k}^2)+\gamma /({\rho }_0{\omega }^2)$ with the outgoing branch ${\gamma }^2=k^2-{\omega }^2/c_0^2$.

For $Q_T+Q_{XXX}+a{Q}_{XX}+b{Q}_X+cQ=0$, an exponential gauge removes the second derivative. With ${\kappa }^2=a^2/3-b>0$ and $\sigma =-c+ab/3-2a^3/27$, the remaining equation is $q_t+q_{xxx}-q_x=0$. Negative $a$ improves spatial decay of the transformed initial data; positive $a$ can destroy the convergence of its Half-range Fourier transform.

For the advection-diffusion equation $u_t=u_{xx}+\alpha u_x$ on $x>0$, the spectral divergence form uses $\omega =k^2-i\alpha k$ and factor $e^{-ikx+\omega t}$. With negative-exponential Half-range Fourier transforms, integration gives the displayed global relation for $\mathrm{Im}k\le 0$. The boundary transform $G_1$ refers to $u_x(0,t)$, the negative of the outward normal derivative.

For the linear dispersive Stokes equation $q_t+q_x+q_{xxx}=0$, let $\omega (k)=i(k-k^3)$. Its local relation is obtained with $X=q_{xx}+ik{q}_x+(1-k^2)q$. Integrating ${\partial }_t(e^{-ikx+\omega t}q)+{\partial }_x(e^{-ikx+\omega t}X)=0$ over the half-line gives the displayed global relation, with negative-exponential Half-range Fourier transforms and $G_j(k,t)={\int }_0^t{e}^{\omega (k)s}{\partial }_x^{j}q(0,s)ds$. The spatial transforms are holomorphic for $\mathrm{Im}k<0$, whereas the finite-time boundary transforms are entire functions.

Use the Half-range Fourier transform and the finite-time spectral boundary transformswhere $g_j={\partial }_x^{j}q(0,\cdot )$ for $j=0,1,2$. The spatial transforms are holomorphic in $\mathrm{Im}k<0$, while these finite-time transforms are entire functions. For now $g_2$ is an unknown boundary trace. Integrating the local relation over the half-strip, using spatial decay at infinity, gives the global relationThe boundary sign follows from integrating the flux derivative: its contribution is minus the flux at zero.

Fourier inversion first gives the initial real-line integral plus a real-line integral of $F$. To put the boundary contribution on its useful complex contour, defineOrient its full boundary with $D_{+}$ on the left. It consists of the real axis from left to right and the upper curve $C:k=u+i\sqrt{1+3u^2}$ from right to left. The real-axis piece must not be discarded for this dispersion. The contour integration representation isTo justify the replacement, the integral of $e^{ikx-wt}F$ along $C$ is zero by closing above $C$. In that region $\mathrm{Re}w\ge 0$, and each boundary term is an integral of $e^{-w(t-s)}$ over $0\le s\le t$ times a polynomial in $k$. This has no growing temporal exponential there; $e^{ikx}$ suppresses the upper closing arc for $x>0$. The Cauchy integral theorem then applies. The figure shows the full contour and, separately, the sectors used in the nonlinear problem.

One may replace $F(k,t)$ by $F(k,T)$ for any finite $T\ge t$. Their difference is a future-time integral over $t<s<T$; its factor $e^{w(s-t)}$ decays inside $D_{+}$, so its integral around $\partial D_{+}$ vanishes. All contour formulas are understood as limits of truncated contours, with standard small displacements if a boundary transform needs an Abel limit. Compatibility of the initial and boundary data controls the corner behavior. This is the two-trace half-line cubic dispersion representation.

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.

Set $\kappa =(a^2/3-b{)}^{1/2}>0$, $\sigma =-c+ab/3-2a^3/27$, and use the gauge reduction of a third-order half-line equationThe spatial differential operator acting on $q$ is obtained by replacing ${\partial }_X$ with $\kappa {\partial }_x-a/3$. Its second-derivative coefficient vanishes. The first-derivative coefficient is $\kappa (b-a^2/3)=-{\kappa }^3$, and the constant coefficient is $c-ab/3+2a^3/27=-\sigma$. The time derivative supplies $\sigma q+{\kappa }^3{q}_t$, so all constant terms cancel and the remaining equation is ${\kappa }^3(q_t+q_{xxx}-q_x)=0$.

The transformed initial condition and Dirichlet boundary data are thereforeCompatibility at the corner is preserved. Since $a<0$, this exponential gauge improves the spatial decay of the initial data, which supports convergence of its Half-range Fourier transform.

Use the negative-exponential Half-range Fourier transformFor the boundary traces $g_j(t)={\partial }_x^{j}q(0,t)$, define finite-time spectral boundary transforms $G_j(k,t)={\int }_0^t{e}^{w(k)s}{g}_j(s)ds$. Integrating the conservation law in $x$ and then time gives the global relationThe plus sign of the boundary contribution follows because the integrated flux is $-E(0,t)X(0,t,k)$.

Fourier inversion consequently gives a real-line initial integral plus a boundary integral. To put the latter in the complex plane, defineOrient $\partial D_{+}$ from the left upper infinity, through $i$, to the right upper infinity, with $D_{+}$ on the left. Each $G_j$ is entire, and in the region between the real axis and this contour, $\mathrm{Re}w\ge 0$. The causal factors $e^{-w(t-s)}$ are bounded there and $e^{ikx}$ decays for $x>0$. The Cauchy integral theorem, with the usual large-arc limiting argument, deforms the boundary integral to giveInitial smoothness, spatial decay and corner compatibility justify inversion; if necessary the integrals are first regularized before contour deformation. The unknown Neumann-type traces $g_1,g_2$ will be removed by the global relation.

The algebraic change of variables still works for $a>0$, but the transformed initial data are now multiplied by $e^{aX/3}$. Mere decay of $Q_0$ does not ensure existence of the Half-range Fourier transform used above. For example, $Q_0(X)=e^{-dX}$ with $0<d<a/3$ gives a growing transformed profile.

There is also a genuine stability distinction. On the whole line a real-frequency mode of the original equation evolves with factorIts modulus is $e^{(a{K}^2-c)T}$, so positive $a$ amplifies arbitrarily high frequencies rather than damping them. Thus the formal gauge equivalence does not justify the same unweighted Sobolev well-posedness conclusion. Weighted or more strongly decaying data and corresponding contour changes may allow particular solutions; the assumptions stated for the negative-$a$ construction cannot simply be reused.

Extend $q$ by zero to the negative half-line. Its ordinary Fourier transform is then exactly the Half-range Fourier transform $\hat{q}(k)$. The Fourier inversion theorem gives the real-line term for $x>0$; no endpoint convention at $x=0$ is needed.

It remains to show that each extra contour contributes zero, independently of its coefficient. For $\mathrm{Im}k<0$, the integral defining $\hat{q}$ is holomorphic. With $\alpha =e^{2\pi i/3}$, multiplication by ${\alpha }^2$ rotates the sector $E$ into $-2\pi /3\le \arg ({\alpha }^{2}k)\le -\pi /3$, and multiplication by $\alpha$ rotates $D$ into $-2\pi /3\le \arg (\alpha k)\le -\pi /3$. Thus the two rotated transforms are holomorphic on their respective upper sectors.

Orient the contours as in the PDF: $\partial E$ runs from infinity on the $\pi /3$ ray into zero and then out along the positive real axis; $\partial D$ runs from negative real infinity into zero and then out along the $2\pi /3$ ray. Both have their sector on their left. Smooth decay and integration by parts give $\hat{q}(k)=O(1/k)$ in closed lower-half-plane sectors. The factor $e^{ikx}$ decays in the upper half-plane for $x>0$. Close each sector by a large arc; the arc integral vanishes by Jordan lemma, including its short portion near the real axis. The Cauchy integral theorem therefore gives the rotated null contours for half-range Fourier inversion:Adding arbitrary constant multiples of these two zero integrals to ordinary Fourier inversion proves the asserted inversion formula. The constants $c_1,c_2$ are arbitrary in this part; selecting them later is what removes an unknown boundary trace.

Define the initial Half-range Fourier transform $Q(k)={\int }_0^{\infty }{e}^{-ikx}{q}_0(x)dx$, and, for each available boundary trace, its finite-time spectral boundary transformTemporarily let $f_j={\partial }_x^{j}q(0,t)$ also denote the missing trace. Three integrations by parts in $x$ giveConsequently the backward-sign Airy half-line global relation isThe spatial transform is holomorphic below the real axis, while each finite-time $F_j$ is an entire function. Moreover, $F_j(\alpha k,t)=F_j({\alpha }^{2}k,t)=F_j(k,t)$ because ${\alpha }^3=1$.

Suppose the missing derivative has order $j$, and put $r=2-j$. Its contribution to the right-hand side is $a_j{k}^r{F}_j$, where $a_0=1$, $a_1=-i$, $a_2=-1$. Apply the inversion from the preceding part at time $t$ and selectThe rotated coefficients of that missing trace now equal minus its real-line coefficient. Its entire contribution is thereforeThe real-axis pieces cancel. What remains is the boundary of the middle upper sector, outward on $\arg k=\pi /3$ and inward on $\arg k=2\pi /3$. In this sector $\mathrm{Im}{k}^3\le 0$, andis bounded by ${\int }_0^t|f_j(s)|ds$. The factor $e^{ikx}$ decays exponentially throughout the closing arc, overcoming $k^r$. There are no poles. Thus the Cauchy integral theorem proves the cubic-dispersion elimination of one missing boundary trace and removes every occurrence of the unknown $f_j$.

For a compact explicit answer, use only the two prescribed traces to form $S(k,t)=Q(k)+H(k,t)$. The three choices areFor each row the solution isAll quantities in this formula are transforms of the given data. On the contour rays $k^3$ is real, so the time exponential is oscillatory; the integrals use the usual limiting interpretation of Fourier inversion. The decay estimate used for elimination belongs to the middle sector, not to $D$ or $E$, where $e^{-i{k}^3(t-s)}$ can grow.

At $t=0$, $F_j=0$ and the preceding inversion recovers $q_0$. The spectral exponential satisfies the Airy equation with the required minus sign. Differentiating the finite-time amplitudes produces only derivatives of the Dirac delta supported at $x=0$ and zero rotated closed-contour contributions, so the interior equation also holds. Finally, two solutions with the same data have $Q=H=0$ for their difference, so this representation gives zero: each of the three prescribed pairs determines the solution uniquely in the smooth decaying class.

Let $G_0(k),G_l(k)$ denote the Half-range Fourier transforms of $g_0,g_l$, and let $N_0(k),N_l(k)$ denote those of the upward derivatives $n_0,n_l$. Also setIntegrate the tangential derivatives in the three spectral functions by parts. For example, ${\int }_0^{\infty }{e}^{-ikx}{g}_0^{\prime }=-h(0)+ik{G}_0$, and ${\int }_0^l{e}^{ky}{h}^{\prime }=e^{kl}h(l)-h(0)-kH$. All corner values cancel in ${\rho }_L+{\rho }_B+{\rho }_T=0$. The result is the semistrip Laplace spectral global relation

For $\kappa >0$, use the sine transformsReality of $q$ makes $W(\kappa )$ and $W(-\kappa )$ real, so taking imaginary parts removes the unknown left-side normal derivative. At $k=\kappa$ and $k=-\kappa$ the two consequences of the global relation areSubtracting these equations and then substituting back gives the semistrip Dirichlet-to-Neumann sine transforms entirely in terms of the prescribed Dirichlet boundary data:These are the requested transforms of $q_y$ itself. The bottom outward Neumann boundary data have the opposite sign. The apparent division at $\kappa =0$ is handled by a continuous limit whenever the corresponding moments exist; there is no singularity for any $\kappa >0$.

Let $s_j=\sin {\beta }_j$, $c_j=\cos {\beta }_j$. Inverting the two orthogonal derivative systems gives, on the vertical axis,and, on the horizontal axis,Using the Wirtinger derivative $q_z=(q_x-i{q}_y)/2$, these become

Define the appropriate boundary transforms byThe first pair is holomorphic for $\mathrm{Re}k<0$; the second pair consists of Half-range Fourier transforms, holomorphic for $\mathrm{Im}k<0$. Parametrizing the vertical boundary by $z=iy$ gives $dz=idy$, which must be retained. Substitution yields the quarter-plane oblique boundary spectral functions:Their signs follow from the stated clockwise boundary orientations: upward on the left and inward on the bottom.

Set $\omega (k)=k^2-i\alpha k$ and $\nu (k)=i\alpha -k$, so $\omega (\nu )=\omega (k)$. For boundary traces $h_j(s)={\partial }_x^{j}q(0,s)$, defineThe Half-range Fourier transform of the equation is obtained by two integrations by parts:Solving this scalar equation gives the drift-diffusion half-line global relationIn particular the reflected relation is valid on $L$, since $\mathrm{Im}\nu (k)\le 0$ there.

In the region between the real axis and $L$, $\mathrm{Re}\omega \ge 0$. Thus a boundary contribution $e^{-\omega t}{G}_j={\int }_0^t{e}^{-\omega (t-s)}{h}_j(s)ds$ admits an upward contour deformation from the real axis to $L$. Large-contour estimates follow from the spatial exponential and the time integral; at the upper time endpoint one can integrate in time once or use a small contour displacement. The initial-data transform is retained on the real axis. Hence, if $I_C[F]=(2\pi {)}^{-1}{\int }_C{e}^{ikx-\omega (k)t}F(k)dk$, ordinary inversion first givesFor the Dirichlet boundary condition, take $h_0=g_0$. Subtract the null integral of the reflected time-dependent transform, using the inversion from part (a) with $c=-1$. The unknown derivative transform cancels because $i\nu +\alpha =-ik$. This derives the Dirichlet spectral formula for half-line constant drift:Only initial and prescribed boundary data occur in this expression.

For the Neumann boundary condition, take $h_1=g_1$. The reflected global relation reads $e^{\omega t}\hat{q}(\nu ,t)=Q(\nu )-G_1+ik{G}_0$. Solve for $G_0$ and substitute it in the preliminary representation. The leftover term is a null integral with multiplier $(ik+\alpha )/(ik)$, analytic above $L$; its pole at zero is below the contour. Thus the Neumann spectral formula for half-line constant drift isThere is no residue from the origin, since this elimination never moves the contour across it. The given derivative is the inward-coordinate derivative $q_x$, not the outward derivative $-q_x$. The initial compatibility for this case is $q_0^{\prime }(0)=g_1(0)$.

Define the negative-exponential Half-range Fourier transformswhich are holomorphic for $\mathrm{Im}k<0$. Write $g_j(s)={\partial }_x^{j}q(0,s)$ and define the finite-time spectral boundary transformsIntegrate the local relation over $0<x<\infty$, $0<s<t$. The spatial boundary term at infinity vanishes by decay, and the term at $x=0$ is subtracted. Consequently the half-line linear dispersive Stokes global relation isFourier inversion on the real axis gives an initial-data integral plus an integral of $B$. To put the boundary term on a complex contour, setFor $k=u+iv$, $\mathrm{Re}\omega =v(3u^2-v^2-1)$, so $D_{+}$ is the connected upper domain $v>0$, $v^2>3u^2-1$. Orient $\partial D_{+}$ with this domain on the left: from upper-left infinity down to $-1/\sqrt{3}$, along the real segment to $1/\sqrt{3}$, and then to upper-right infinity.