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For a barotropic equation of state, $\delta P=c_s^2\delta \rho =c_s^2\bar{\rho }\delta$. A spatial Fourier mode of comoving wavenumber $k$ therefore satisfiesThe Jeans wavenumber is defined by equality of the pressure and self-gravity terms:For $k\gg k_J$, pressure dominates and produces acoustic oscillations whose amplitude is affected by Hubble friction. For $k\ll k_J$, self-gravity dominates and a growing mode develops: this is Jeans instability.

For $\mathbf{u}=\nabla \times (\psi \hat{\mathbf{y}})$,Hence $u_x+w_z=-{\psi }_{zx}+{\psi }_{xz}=0$, so mass conservation holds identically. Removing the hydrostatic part by writingand taking the curl of the Stokes flow equation gives the biharmonic equation

For a Fourier mode proportional to $e^{ikx}$ with $k>0$, decay as $z\to -\infty$ selectsThe velocity and pressure fields are thereforeDirect substitution verifies the momentum equation.

For separation of variables, set $u=X(x)Y(y)$, with $X^{\prime \prime }/X=-Y^{\prime \prime }/Y=-\lambda$. The side Neumann boundary conditions give $X^{\prime }(0)=X^{\prime }(\pi )=0$. As in the standard Sturm-Liouville problem, integration by parts shows $\lambda \ge 0$. The zero eigenvalue gives constant $X$, and $Y^{\prime \prime }=0$ together with $Y^{\prime }(\pi )=0$ makes $Y$ constant too.

For positive eigenvalues, the side conditions give $\lambda =n^2$, $X_n=\cos (nx)$, $n\ge 1$. The top condition chooses the hyperbolic cosine factor $\cosh (n(\pi -y))$. Normalize it to one at $y=0$, obtainingThe orthogonality of integer Fourier modes and the permitted interchange of summation and integration giveFor the remaining Fourier coefficients, integrate by parts twice. The sine boundary term vanishes, and with $f=x(\pi -x)$,The last integral vanishes because $f^{\prime \prime }=-2$ and $n$ is a positive integer. Thus $a_n=-2(1+(-1{)}^n)/n^2$: only even Fourier modes survive, as expected from reflection symmetry about $x=\pi /2$. Writing $n=2m$ givesThe factor involving hyperbolic cosine lies between zero and one on the closed rectangle. Comparison with $\sum m^{-2}$ proves uniform convergence, so the solution is continuous on the entire boundary and takes the prescribed bottom values. On every strip $y\ge \epsilon >0$, the Fourier coefficients have additional exponential decay, allowing differentiation and verifying Laplace's equation and all three Neumann boundary conditions.

To justify the endpoint calculation in b, differentiate at $0<x<\pi$, $y=0$ in the interior-edge sense:Subtracting $1$ from the hyperbolic tangent leaves an absolutely convergent series. For the remaining part, the power series $-\log (1-z)={\sum }_{m\ge 1}{z}^m/m$ gives, by taking real parts at $z=r{e}^{2ix}$ and then $r\uparrow 1$,The same formula follows as the limit of the differentiated interior solution, since its leading multiplier is $e^{-2my}$. This establishes the boundary normal derivative used in b. It also explains why the failure of the corner compatibility condition for mixed boundary data affects first derivatives at the bottom corners without spoiling continuity of $u$ itself.

The trigonometric identityshows that only angular Fourier modes $1$ and $3$ are required. Differentiating the harmonic expansion on a disk at $r=1$ multiplies the coefficient of each nonconstant Fourier mode by $n$. Thus $B_1=3/4$ and $3B_3=-1/4$, givingEach term is a regular harmonic function, and the radial derivative isThe Neumann compatibility condition for Laplace's equation holds becauseas is necessary by the divergence theorem applied to the gradient of a harmonic function.

The additive constant is undetermined. This Neumann boundary condition is unchanged by $T\mapsto T+C$, so the paper's blanket uniqueness assertion requires a normalization in this case. The uniqueness of the Neumann problem gives uniqueness up to a constant; prescribing $T(0)=0$, or zero mean temperature, selects the displayed answer with $C=0$.

The boundary data consist of the third angular Fourier mode. Its regular radial factor in the harmonic expansion on a disk is $r^3$, soFor a direct check of Laplace's equation,Also $T(1,\theta )=\sin (3\theta )$. At the origin the apparent coordinate singularity is harmless: this is the harmonic polynomial $3x^{2}y-y^3$. The allowed uniqueness for the Dirichlet problem completes the identification.

Separation of variables gives $X^{\prime \prime }=-k^{2}X$ and $T^{\prime }=-\kappa k^{2}T$. The periodic matching of both the function and its first derivative selects $k=n\pi /L$, with the constant mode at $n=0$. Thus the periodic diffusion on a circle solution iswhere the Fourier coefficients of the initial function areThe formula for $a_n$ also gives $a_0$ at $n=0$. The constant mode preserves the initial spatial mean; every nonconstant Fourier mode decays. For initial data in L2 space, the series approaches those data in L2 space as $t\downarrow 0$ and is smooth at positive times.

The only nonzero angular Fourier mode is $\cos (2\sigma )$. Its radial equation is $T^{\prime \prime }-4T=0$. A solution vanishing at $\tau =a$ is proportional to $\sinh [2(\tau -a)]$; making its value one at $b$ fixes the constant. The Dirichlet problem between bipolar coordinate circles therefore givesIts second derivatives in $\sigma$ and $\tau$ cancel, and the endpoint values are respectively zero and $\cos (2\sigma )$. The difference of any two solutions would be a harmonic function with zero Dirichlet boundary conditions, so the maximum principle proves uniqueness.

Geometric wording. In the printed bipolar coordinates, the circle radius $1/\sinh \tau$ decreases with $\tau$. For $0<a<b$, the $b$ circle is contained inside the $a$ circle. Thus the source’s “inner” and “outer” labels are interchanged geometrically. The solution above follows the explicit coordinate boundary values, which unambiguously specify the problem.

Let $d=b-a$. In the mixed Dirichlet-Neumann problem in an elliptic annulus, the angular Fourier modes separate. The outer data excite the $\cos \nu$ mode, while the inner derivative excites $\cos 2\nu$.

For the first mode, a radial factor with zero derivative at $\mu =a$ is $\cosh (\mu -a)$. Normalizing its outer value gives $F_1(\mu )=\cosh (\mu -a)/\cosh d$. For the second mode, a radial factor vanishing at $\mu =b$ is $\sinh [2(b-\mu )]$. Its derivative at $a$ must be $-1$, fixing the coefficient to $1/[2\cosh (2d)]$. ThereforeEach term satisfies Laplace's equation; at $\mu =b$ the second term vanishes and the first is $\cos \nu$. At $\mu =a$, differentiating gives $0-\cos 2\nu$, as required. All other modes, including the constant mode, vanish by the homogeneous mixed conditions. Alternatively, integration by parts applied to the difference of two solutions gives a zero integral of $|\nabla \Phi {|}^2$, proving uniqueness.

Boundary convention. The calculation follows the explicitly printed coordinate derivative ${\partial }_{\mu }\Phi$. In elliptic coordinates, the scale factor is $h=\sqrt{{\sinh }^2\mu +{\sin }^2\nu }$ and the outward unit normal at the inner boundary points toward decreasing $\mu$. A physical normal derivative there would be $-h^{-1}{\partial }_{\mu }\Phi$. Thus the phrase “normal derivative” in the source must be understood in its displayed coordinate sense; imposing the same numerical data on a unit-normal derivative would be a different problem.

For separation of variables matching the boundary Fourier mode, take ${\psi }_0=X(x)e^{iky}$ with $X(0)=1$. Because ${\partial }_y^3{e}^{iky}=-i{k}^3{e}^{iky}$, the reduced ordinary differential equation isThereforeThe decay of a Fourier mode in a third-order spatial evolution equation requires a sign check: this satisfies the requested limit at positive infinity precisely when $\boxed{k>0}$. For $k=0$ it remains identically one, while for $k<0$ its amplitude grows exponentially. Hence for those nonpositive values there is no separated solution meeting both prescribed boundary conditions; the printed assumption that $k$ is merely real needs this additional restriction for decay.

Treat each angular Fourier mode independently. Boundedness at the axis excludes the negative radial powers and the logarithm inside; decay at infinity excludes the positive powers, constant and logarithm outside. For the cosine mode writeContinuity at $R$ gives $B_n=A_n{R}^{2n}$. The jump of the normal derivative is thenThus $A_n=a_n{R}^{1-n}/(2n)$ and $B_n=a_n{R}^{n+1}/(2n)$. The sine mode has the same relation with $b_n$. Therefore the potential of a charged cylindrical shell with angular Fourier data isThere is no angular mean term: the given surface charge density has zero average. The interior constant is zero because its boundary value must match the vanishing exterior mean. A nonzero mean charge per unit length would instead produce a logarithmic electrostatic potential, incompatible with the specified decay at infinity.

Reflecting the solution across $\theta =\pi \alpha$ preserves its Dirichlet boundary conditions and outer boundary data, so uniqueness makes the solution reflection-symmetric. Sinceonly odd angular Fourier modes survive. Put $\rho =(r/R{)}^{1/(2\alpha )}$ and $\psi =\theta /(2\alpha )$. On the symmetry axis $\psi =\pi /2$, comparison of the convergent power series givesThusThis is the imaginary part of $\sin (\rho e^{i\psi })$, so a useful closed form isAt either straight edge $\sin \psi =0$, while on the symmetry axis the formula is $\sinh \rho$. Every term is harmonic in the sector, and the factorial denominator gives convergence up to $r=R$, verifying the full solution.

When $a=0$, the explicit amplification factor becomesA single Fourier mode has constant modulus in space and is not in the l2 sequence space. Instead choose a whole interval of frequencies on which there is definite growth. LetThe inverse Fourier transform here gives the real sequenceBy the Parseval identity, its squared norm is $\Vert U^0{\Vert }_{{\ell }_2}^2=\Delta x|E|/(2\pi )=\Delta x/2$, finite and strictly positive. These data are also compatible with sampling a bounded continuous function: useThe value at zero is the limit of the displayed quotient, and $u_0(j\Delta x)=U_j^0$.

On $E$ we have ${\sin }^2\theta \ge 1/2$. Therefore the transformed solution and the Parseval identity giveFor each fixed pair $\Delta x,\Delta t>0$, takeTaking square roots proves $\Vert U^m{\Vert }_{{\ell }_2}\ge (1+c{)}^m\Vert U^0{\Vert }_{{\ell }_2}$ for every $m\ge 1$. In particular the norm already increases at the first step, so no positive mesh pair gives practical stability in the contraction sense of (b).

The native PDF places the existence of $c>0$ before the universal quantifier over mesh pairs. If this means a single positive $c$ independent of $\Delta x,\Delta t$, that assertion is false for nonzero data. Indeed, for every datum the same Fourier symbol givesFor any proposed fixed $c>0$, choose $0<\nu <\sqrt{(1+c{)}^2-1}$. The upper bound then contradicts the requested lower bound at $m=1$. The correct quantifiers are $\forall \Delta x,\Delta t>0\exists c=c(\Delta x,\Delta t)>0$, with nonzero square-summable data as constructed above.

Finally, failure of this stronger practical stability of a finite difference scheme should not be confused with every possible refinement notion of numerical stability. If $\Delta t/(\Delta x{)}^2\le K$, then $m\Delta t\le T$ givesThus a finite-time mesh-uniform bound can hold under this severe refinement even though contraction fails on every positive mesh. This is the distinction recorded in centered-advection refinement-dependent stability.

Modes, stability and symmetry. A normal mode of the linearized system has a fixed displacement pattern multiplied by one temporal oscillation. For $n$ independent generalized coordinates, a conservative small-disturbance system hasWhen the mass matrix $\mathsf{M}$ is positive definite and the stiffness $\mathsf{K}$ is a symmetric matrix, the generalized eigenvalue problem for small oscillations has $n$ independent mode directions, counting degeneracy. Positive ${\omega }^2$ gives stable oscillation, negative ${\omega }^2$ an exponentially growing and decaying pair, and zero ${\omega }^2$ a neutral direction whose nonlinear stability needs further analysis. An arbitrary linear motion is a superposition of these modes.

If a symmetry preserves both kinetic and potential energies, it maps modes to modes of the same frequency. Mode patterns can be classified by that symmetry, for example reflection-even and reflection-odd patterns. Degenerate eigenspaces admit different bases, so a mode shape need not be uniquely selected inside such a subspace.

Two pendulums free to swing in any direction. Let $x_j$ be the small horizontal displacement along the line joining the supports and $y_j$ the perpendicular horizontal displacement. Put ${\omega }_0^2=g/L$ and $\kappa =k/m$. To quadratic order,The spring is unstretched at equilibrium. Its first-order length change is only $x_2-x_1$; a transverse difference changes its length at second order and produces no quadratic transverse restoring energy. Consequently the four modes areThis is the transverse degeneracy of unstretched spring-coupled pendulums: ${\omega }_0$ has a three-dimensional eigenspace. Independent transverse motions form an equally valid basis for two of those directions. A pre-tensioned spring would change the transverse result.

The open three-pendulum chain. Restrict motion to the vertical plane containing the supports, with horizontal coordinates $x_1,x_2,x_3$. There are springs between neighbours and no wall springs. The equations areThe spring matrix has eigenvalues $0,1,3$, with respective eigenvectors $(1,1,1)$, $(1,0,-1)$ and $(1,-2,1)$. The normal modes of three freely ended coupled pendulums are thereforeThe lowest mode stretches neither spring. In the middle mode the outer masses oppose each other and the central mass is stationary. The highest mode has the central mass moving oppositely with twice the outer amplitude. Reflection about the middle support makes the first and last patterns even and the middle pattern odd.

An even ring. For a large ring, use tangential displacement $x_j$ and periodic indexing $x_{j+N}=x_j$. The locally straight nearest-neighbour equations areCyclic translation symmetry selects Fourier modes $x_j\propto e^{i(qj-\omega t)}$, with $q=2\pi s/N$. HenceFor even $N$, the lowest pattern has equal tangential displacements, while the highest alternates signs around the ring. Thus

These are the even and odd pendulum-ring modes. For exact finite regular-polygon geometry with straight chord springs, the first-order spring extension includes a factor $\cos (\pi /N)$, so its squared contribution is multiplied by ${\cos }^2(\pi /N)$. That correction tends to one in the requested large-$N$ limit.

An odd ring. Perfect alternation cannot close after an odd number of steps. The two highest wave indices are instead $s=(N-1)/2$ and $(N+1)/2$. They give a degenerate pair, represented by real sine and cosine standing patterns, withThe maximum is slightly below the even-ring limiting value, and the sign pattern must contain a gradual phase mismatch rather than exact alternation. The uniform lowest mode is unchanged, and there are still $N$ independent tangential mode directions.

Put $K=(k^2+m^2{)}^{1/2}$ and $P=p^{\prime }/{\rho }_0$. Acting on the Fourier mode, $D_0=-i(\hat{\omega }-Uk)=-i\omega$. Continuity and horizontal momentum give $u=-mw/k$ and $P=\omega u/k$. The buoyancy equation gives $g{\rho }^{\prime }/{\rho }_0=i{N}^{2}w/\omega$. Substitution into vertical momentum yieldsThe two signs describe the two frequency branches; when the intrinsic frequency is required positive, $\omega =N|k|/K$. For a stationary disturbance $\hat{\omega }=0$, so $\omega =-Uk>0$. With $U>0$ this requires $k<0$ and selects the minus sign, $\hat{\omega }=Uk-Nk/K$. The stationary dispersion condition isDifferentiate the unconstrained dispersion before imposing stationarity. The stationary mountain-wave group velocity is obtained from the ground-frame dispersion; before imposing stationarity it isOn the stationary locus this becomesDifferentiating after substituting $K=N/U$ as though all neighbouring modes were stationary would incorrectly erase the frequency gradients.

For upward energy propagation from the lower boundary, choose $m<0$, since $k<0$ and ${\hat{c}}_{gz}>0$. It requires real nonzero $m$, henceThe downstream ray inclination above horizontal isIt is fixed for a given monochromatic forcing, with the two signs of $m$ giving the upward/downward rays.

Frame qualification for the requested phase angle: the normal intrinsic phase velocity is ${\mathbf{c}}_p=\omega (k,m)/K^2$. Because $\omega =-Uk$, the boxed stationary group velocity is exactly $-{\mathbf{c}}_p$: its angle to intrinsic phase propagation is $\pi$, and it is perpendicular to the stationary phase lines. By contrast the intrinsic group velocity ${\mathbf{c}}_g=\widehat{{\mathbf{c}}_g}-U\hat{\mathbf{x}}$ satisfies ${\mathbf{c}}_g\cdot (k,m)=0$, so its angle to intrinsic phase propagation is $\pi /2$. In the ground frame the phase velocity of a strictly stationary pattern is zero and has no nonzero propagation direction. For example $k=-0.6N/U$, $m=-0.8N/U$ gives $\widehat{{\mathbf{c}}_g}=(0.36U,0.48U)$ and ${\mathbf{c}}_p=(-0.36U,-0.48U)$. These distinctions prevent applying the usual zero-flow perpendicularity rule to two velocities expressed in different frames.

For an error Fourier mode $e_j^n={\hat{e}}^n{e}^{i\theta j}$, the periodic discrete Laplacian has eigenvalue $-4{\sin }^2(\theta /2)/h^2$. The Backward Euler method therefore has amplification factorvon Neumann stability analysis thus imposes no upper limit on positive $\Delta \tau$. The constant mode has $g=1$ and is not damped; compatible forcing and a gauge condition remain necessary as discussed in part (e). Large steps can accelerate relaxation of nonconstant modes, but require a suitably accurate linear solver and can worsen its conditioning. Unconditional linear stability is not a statement about time accuracy or cost.

Write the membrane height above its mean plane as a sum of Fourier modes, $h(\mathbf{r})={\sum }_{\mathbf{q}}{h}_{\mathbf{q}}{e}^{i\mathbf{q}\cdot \mathbf{r}}$. The supplied average is a thermal mean-squared modal height; for complex Fourier coefficients it is understood as $⟨|h_{\mathbf{q}}{|}^2⟩$, with $h_{-\mathbf{q}}=h_{\mathbf{q}}^{\ast }$ for a real height field. The terms are:

In the small-slope approximation, Helfrich membrane energy containsThe equipartition theorem, with the independent real modes counted consistently, gives the stated membrane fluctuation spectrum under tension. Its units are length squared: $\sigma q^2$ and $\kappa q^4$ both have units $\mathrm{J}{\mathrm{m}}^{-4}$, so $k_{B}T/[A(\sigma q^2+\kappa q^4)]$ has units ${\mathrm{m}}^2$.

At small length scales, $q$ is large and the bending term dominates:The fluctuation amplitude is controlled mainly by the membrane bending modulus and thermal energy, rather than tension. The crossover occurs when $\sigma q_c^2=\kappa q_c^4$, henceThe bending-dominated regime extends up to the membrane elastic length when scale means $1/q$, or up to ${\lambda }_c$ when it means wavelength. At still larger scales tension gives a $q^{-2}$ spectrum. At the opposite end, the continuum formula requires wavelengths larger than the microscopic membrane thickness or molecular cutoff. Without numerical $\kappa$ and $\sigma$, no numerical crossover scale is determined.

Two-dimensional incompressibility permits $u={\psi }_y$, $v=-{\psi }_x$. Taking the curl of the Stokes momentum equation eliminates pressure and gives ${\nabla }^2(v_x-u_y)=0$. Since $v_x-u_y=-{\nabla }^2\psi$, the stream function obeys $\boxed{{\nabla }^4\psi =0}$.

Scale radius by $a$, velocity by $U$ and stream function by $Ua$. In polar coordinates $u_r=r^{-1}{\psi }_{\theta }$, $u_{\theta }=-{\psi }_r$. No penetration makes $\psi$ constant on the cylinder; choose that constant zero. No slip also requires ${\psi }_r=0$ at $r=1$. The uniform stream has $\psi =r\sin \theta$, specifying the far-field condition.

The boundary forcing has angular mode $\sin \theta$, so its corresponding mode must have $\psi =f(r)\sin \theta$. Define $L_1=d^2/d{r}^2+r^{-1}d/dr-r^{-2}$. Then ${\nabla }^4\psi =(L_1^{2}f)\sin \theta$, while $L_1{r}^s=(s^2-1)r^{s-2}$. Thus the indicial equation of $L_1^2$ is $(s^2-1)[(s-2{)}^2-1]=(s-3)(s-1{)}^2(s+1)=0$, givingThe repeated exponent one supplies the logarithmic term. The condition $f^{\prime }\to 1$ first forces $A=B=0$ and $C=1$. Then $f(1)=0$ forces $D=-1$, but $f^{\prime }(1)=C-D=2\ne 0$. No additional Fourier mode can repair the missing boundary condition in this angular mode. Therefore the unbounded-domain Stokes problem has no solution with both no slip and a nonzero uniform far stream.

This is Stokes' paradox. It identifies a nonuniform approximation, rather than the absence of physical flow. For any small but nonzero Reynolds number, convective inertia becomes important sufficiently far away. An inner Stokes region must be matched to an outer Oseen or Navier-Stokes region that retains that inertia. Alternatively, finite external boundaries introduce a different well-posed physical problem; they cannot be sent to infinity while keeping the naive Stokes solution unchanged.

Use the forward spatial transform $\hat{f}(k)=\int e^{-ikx}f(x)dx$ and inverse $f(x)=\int e^{ikx}\hat{f}(k)dk/(2\pi )$. The retarded transformed Green function obeysIt vanishes for negative $\tau$. Continuity at zero excludes an unwanted ${\delta }^{\prime }$ term; integrating across zero then fixes $G_k^{\prime }(0^{+})=c^2$. Define $\gamma =\alpha c^2$ and $d_k=\sqrt{{\gamma }^2-c^2{k}^2}$. The two homogeneous exponents are $-\gamma \pm d_k$. Imposing the two matching conditions gives the Fourier-mode Green function of the telegraph equation,For an explicit time-Fourier route, use $G_k(\tau )=\int e^{-i\omega \tau }{\mathcal{G}}_k(\omega )d\omega /(2\pi )$. ThenFor $k\ne 0$, both poles lie below the real axis. Closing above for $\tau <0$ gives zero; closing below for $\tau >0$ is clockwise, giving $-i$ times the residue sum. Since ${\mathcal{G}}_k=-c^2/[(\omega -{\omega }_{+})(\omega -{\omega }_{-})]$, this iswhich reduces to the same damped sine or hyperbolic sine above. The $k=0$ pole on the real axis is interpreted by the retarded limiting prescription, or directly by the initial-value matching above.

For $|k|<\alpha c$ the response is overdamped and nonoscillatory. For $|k|>\alpha c$, put ${\Omega }_k=\sqrt{c^2{k}^2-{\gamma }^2}$; analytic continuation gives $G_k=\Theta (\tau )c^2{e}^{-\gamma \tau }\sin ({\Omega }_k\tau )/{\Omega }_k$, a real damped oscillation. At equality the finite limit is $\Theta (\tau )c^2\tau e^{-\gamma \tau }$.

All nonzero Fourier modes decay. The uniform mode instead has $G_0=(1-e^{-2\gamma \tau })/(2\alpha )$ for positive time, leaving a constant temperature increment. At small wavenumber the slow decay exponent is $-k^2/(2\alpha )+O(k^4)$, so the diffusion limit of the damped heat-transmission equation has diffusivity $1/(2\alpha )$. The full hyperbolic equation has finite propagation speed $c$, unlike the limiting parabolic heat equation.

Put $R=h_t-\nu {\nabla }^{2}h$, where the diffusion coefficient $\nu$ is constant. ThenOne temporal and two spatial integrations by parts givewith compactly supported variations. HenceHere ${\nabla }^4=({\nabla }^2{)}^2$ is the biharmonic operator. The cross terms cancel because $\nu$ is constant and spatial and temporal derivatives commute. Equivalently, $P^0=2R$ and $Q^{ij}=-2\nu R{\delta }^{ij}$ in part (b), producing $-2{\partial }_{t}R-2\nu {\nabla }^{2}R=0$.

This squared diffusion-residual action does not have only the first-order diffusion equation as its stationary equation. The zero-residual paths $h_t=\nu {\nabla }^{2}h$ have $S=0$ and are certainly solutions, but the variational equation is the higher-order one above. For a spatial Fourier mode with nonzero wavevector $\mathbf{k}$, it givesBoundary data or an additional physical selection principle are needed to retain only the decaying diffusion branch. The spatially uniform mode instead has $h_0(t)=A_0+B_{0}t$ in the stationary equation.

With the PDF's inverse Fourier transform convention, the derivatives act asso its forced differential operator becomes ${\omega }^2-k^2+i\gamma \omega k^2$. ThusFor the physical impulse response choose the retarded fundamental solution, vanishing before $t=0$. For $k\ne 0$ and $\gamma >0$, its poles lie in the lower half-plane. The retarded Fourier pole prescription $\omega \mapsto \omega +i0$ specifies the zero mode and the zero-damping limit as well; a Green function without a temporal or boundary prescription would otherwise permit homogeneous additions.

This is the Kelvin-Voigt rod Green function. Multiplying the homogeneous real-space equation by $-1$ givesA spatial Fourier mode therefore satisfies ${\ddot{\varphi }}_k+\gamma k^2{\dot{\varphi }}_k+k^2{\varphi }_k=0$: positive $\gamma$ really damps the motion. For sufficiently decaying fields its energy satisfiesThis checks the sign of the imaginary term in $\hat{G}$ independently of the transform calculation.