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Scale axial distance with $\lambda \hat{h}$ and normal velocity with $\hat{U}/\lambda$, while the dimensionless axial velocity has scale $\hat{U}$. Let $r=R+y$; keep $R>0$ fixed on the asymptotic scales. In the core, the ratio of transverse viscous terms to axial convective terms is of order $\lambda /Re$. A small indentation and the associated displacement create tangential slip of order $ϵ$ at the edges of the wall layers.

For a layer of relative thickness $\delta \ll ϵ$, that slip makes axial convection of order ${ϵ}^2/\lambda$, while transverse viscous diffusion is of order $ϵ/(Re{\delta }^2)$. Balancing them givesUnsteady diffusion gives the equivalent estimate $\delta \sim (ReSt{)}^{-1/2}$ when $\lambda St=O(ϵ)$. Thus a consistent regime supporting thin viscous boundary layers and an inviscid core isAt a location where both imposed unsteadiness and slip vanish, a shear-dominated entrance balance can instead have thickness $(\lambda /Re{)}^{1/3}$; the same strict Reynolds-number condition keeps even that scale below $ϵ$. The reduction below assumes that no wall-layer thickening invalidates the stipulated small thickness. The normal flux displaced by a slip layer is of order $ϵ\delta$, smaller than the retained ${ϵ}^2$ core flux. Tangential no slip is imposed within these layers, not directly on the leading inviscid core.

Continue the expansion established in part (b), with $p={ϵ}^2{p}_2+\cdots$ after removal of the background pressure. The radial momentum equation at order ${ϵ}^2$ givesIntegrating across the core, then differentiating in $x$, givesThis is the dispersive contribution from transverse inertia.

At second order the axial momentum equation isFor the displacement fields just obtained,Use $u_0=0$ on each wall. The lower and upper regular core limits respectively givewhere $v_2(1)=0$ at the fixed outer wall.

At the moving inner wall $y=ϵF$, no penetration requires $v=ϵ\lambda St{F}_t+ϵu{F}_x$. Expanding about $y=0$, the axial edge velocity is $ϵ{\gamma }_0(F+A/R)+O({ϵ}^2)$ and $v_{1y}(0)=-{\gamma }_0{A}_x/R$. ThusorThe thin wall layers do not change this normal-flux condition at the retained order, because their flux correction is $o({ϵ}^2)$.

Subtract the two axial pressure-gradient limits and use the radial-pressure difference. With ${\alpha }_1={\gamma }_0/R+{\gamma }_1/(R+1)$ and ${\alpha }_2={\gamma }_0^2/R^2-{\gamma }_1^2/(R+1{)}^2$, the result is the forced annular displacement equationEvery coefficient follows from the given profile and wall slopes; no explicit base-flow solution has been used. The derivation is conditional on the regular core limits and wall layers remaining asymptotically thin.

For the reflection of a P-wave from a rigid plane, use the displacement convention $u_x={\varphi }_x+{\psi }_y$, $u_y={\varphi }_y-{\psi }_x$, where $\varphi$ is the P potential and $\psi$ the SV potential. Put $\omega =k{c}_p$, $\kappa =k\sin \theta$, $a=k\cos \theta$, and $b=\sqrt{{\omega }^2/c_s^2-{\kappa }^2}$. The reflected waves must share the tangential wavenumber and frequency and propagate towards negative $y$, so writeThe rigid barrier requires both displacement components to vanish. On $y=0$ this gives $\kappa (1+A)-bB=0$ and $a(1-A)-\kappa B=0$. Solving,The total P potential is the given incident wave plus ${\varphi }_R$, and the S potential is ${\psi }_R$. Reversing the sign convention for $\psi$ reverses $B$ but leaves the displacement unchanged. At normal incidence $A=1,B=0$.

An evanescent reflected S wave would require $b^2<0$, or $\sin \theta >c_p/c_s$. In a stable isotropic elastic solid $c_p>c_s$, so this is impossible for real incidence angles: $b^2=k^2[(c_p/c_s{)}^2-{\sin }^2\theta ]>0$. The reflected S wave is never evanescent in the physical isotropic case. The inequality identifies the hypothetical critical-angle case if the speed ordering were reversed; it is S-to-P conversion, not this P-to-S conversion, that can become evanescent in an ordinary solid.

Let ${\rho }_i={\rho }_n=\rho$ and $\mu ={\mu }_{in}={\mu }_{ni}$. Use normal modes proportional to $e^{i\mathbf{k}\cdot \mathbf{x}-i\omega t}$, so $\mathrm{Im}\omega <0$ means damping. The linear displacements satisfyFor a nonzero-frequency perturbation, linearizing the ideal magnetohydrodynamic induction equation givesProject both momentum equations with $\mathsf{P}=I-\mathbf{k}{\mathbf{k}}^T/k^2$. This removes the pressure gradients. The magnetic tension is already transverse to $\mathbf{k}$, and, with $a=k_{\parallel }{v}_A$, the projected equations areHere $v_A=B_0/\sqrt{4\pi \rho }$ uses the ion density, as stipulated. For either of the two independent transverse polarizations, the equal-density ion-neutral Alfvén dispersion relation follows fromIts determinant is $\omega [{\omega }^3+2i\mu {\omega }^2-a^2\omega -i\mu a^2]$. Removing the static neutral-displacement factor givesThe removed factor is a time-independent neutral displacement with no velocity or magnetic perturbation, a relabeling of the homogeneous neutral fluid rather than a fourth physical wave. Direct elimination in the first-order system for $(u_i,u_n,\delta B)$ gives exactly the same cubic. This also handles the zero-frequency limits without introducing a displacement-label mode.

The neutral displacement equation gives, for each nonzero-frequency root,The vector displacements are parallel as complex polarization vectors, with this scalar amplitude and phase ratio.

For the rapidly damped root, $\mu -i{\omega }_d=-\mu +a^2/(4\mu )+O(a^4/{\mu }^3)$, and henceThe two fluids move almost equally and oppositely; collisions remove the counterflow at a rate close to $2\mu$.

For either weakly damped Alfvén wave, expand the ratio around unity:Equivalently, the exact slippage relation is ${\mathbf{\xi }}_i-{\mathbf{\xi }}_n=-(i\omega /\mu ){\mathbf{\xi }}_n$. Thus the relative slippage has size $|a|/(\sqrt{2}\mu )\ll 1$ and is primarily a small phase difference. The fluids almost share a common displacement, but magnetic tension acts directly on the ions and drag accelerates the neutrals. This imperfect locking causes ambipolar damping.

The sign of the damping is also checked by the perturbation energy balance. With periodic or vanishing-flux boundaries,It follows by dotting each linear momentum equation with its velocity and the induction equation with $\delta \mathbf{B}/(4\pi )$; magnetic-tension work cancels magnetic-energy change, pressure contributes only boundary flux, and the two drag terms combine into the negative square. Even though spatial diffusion was neglected, ion-neutral friction still dissipates the wave energy.

Matter corotating with the footpoint has fixed angular velocity ${\Omega }_0=(GM/R_0^3{)}^{1/2}$. Its effective potential in that rotating frame is the sum of Newtonian gravitational potential and centrifugal potential,Along an outward-inclined straight magnetic field line, write $R=R_0+s\sin i$ and $z=s\cos i$, where $s$ is distance from the footpoint. The exact potential along it isIts first derivative at $s=0$ is zero by the Keplerian rotation balance. At the footpoint the meridional Hessian is $\mathrm{diag}(-3{\Omega }_0^2,{\Omega }_0^2)$ with zero mixed derivative. HenceWhen $0\le i<30^{\circ }$, the potential initially rises and a cold parcel must overcome a barrier. When $i>30^{\circ }$, its quadratic curvature is negative, so the footpoint is a local maximum along the field and a small outward displacement is accelerated further outward:This is the thirty-degree magnetocentrifugal launching criterion. It describes the lack of a local cold-launch barrier, not a nonzero force on a parcel exactly at rest at $s=0$. At exactly $30^{\circ }$ the quadratic term vanishes; for the specified straight line the next term is $-7{\Omega }_0^2{s}^3/(16R_0)$, as in marginal straight-line launch at thirty degrees. Global wind escape additionally depends on conditions farther along the line.

The equilibrium quantities $\rho ,p,\mathbf{B},\Phi$ are real, and the gravitational Hessian ${\Phi }_{,ij}={\partial }_i{\partial }_j\Phi$ is symmetric. The crucial tensor property isThis follows directly from the displayed expression for $V$: the products ${\delta }_{ij}{\delta }_{kl}$ and ${\delta }_{il}{\delta }_{jk}$ are unchanged by interchanging the pairs $(i,j)$ and $(k,l)$; $B_j{B}_l{\delta }_{ik}$ is also unchanged; and the two terms involving $B_i{B}_j{\delta }_{kl}$ and $B_k{B}_l{\delta }_{ij}$ exchange with each other. All coefficients are real.

Multiply $F\mathbf{\xi }$ by $\rho {\mathbf{\eta }}^{\ast }$ and integrate by parts over all space. For admissible displacement fields, the boundary contribution at infinity vanishes: the fluid pressure and density are confined to the body, and the magnetic stress decays in the surrounding conducting medium. The conducting exterior is included in the integration. For a piecewise description, the usual displacement and traction matching conditions cancel the two opposite interface contributions; alternatively, take a smooth transition and then its limit. ThusThe first term is a Hermitian pairing because ${\Phi }_{,ij}={\Phi }_{,ji}$, and the second is Hermitian because $V_{ijkl}=V_{klij}$. Taking the complex conjugate of the same expression with $\mathbf{\eta }$ and $\mathbf{\xi }$ interchanged and relabelling $(i,j)\leftrightarrow (k,l)$ reproduces it. HenceThis establishes the required self-adjoint operator identity in the mass-weighted inner product, on the physical displacement domain with the stated decay and interface conditions. The negligible-density exterior can be treated as a small-positive-density conducting limit when defining this weighted pairing; its magnetic contribution is retained even as its inertia becomes negligible.

First suppose $W[\mathbf{\xi }]>0$ for every nonzero admissible displacement. The conserved energy from part (b) then bounds both $T$ and $W$ by $E$. In particular,where $\Vert \mathbf{\xi }{\Vert }_{\rho }^2=⟨\mathbf{\xi },\mathbf{\xi }⟩$. The positive quadratic form $\Vert \dot{\mathbf{\xi }}{\Vert }_{\rho }^2+2W[\mathbf{\xi }]$ defines the perturbation energy norm and is conserved. Small perturbations in this norm remain small, proving energetic linear stability. Also $\Vert \mathbf{\xi }(t){\Vert }_{\rho }\le \Vert \mathbf{\xi }(0){\Vert }_{\rho }+\sqrt{2E}t$, excluding exponential displacement growth. For any admissible normal mode $\mathbf{\xi }=\mathbf{\zeta }{e}^{-i\omega t}$, multiplication of $-{\omega }^2\mathbf{\zeta }=F\mathbf{\zeta }$ by ${\mathbf{\zeta }}^{\ast }$ gives directlyso all such frequencies are real. No variational principle has been assumed.

To prove instability, suppose some admissible ${\mathbf{\xi }}_0$ has $W_0=W[{\mathbf{\xi }}_0]<0$, and put $N_0=\Vert {\mathbf{\xi }}_0{\Vert }_{\rho }^2>0$. Choose the initial velocityIts kinetic energy exactly cancels $W_0$, so $E=0$ for the entire subsequent motion. Write $N(t)=\Vert \mathbf{\xi }(t){\Vert }_{\rho }^2$. Since $F$ is self-adjoint,Zero energy gives $⟨\mathbf{\xi },F\mathbf{\xi }⟩=\Vert \dot{\mathbf{\xi }}{\Vert }_{\rho }^2$, and hence $N^{\prime \prime }=4\Vert \dot{\mathbf{\xi }}{\Vert }_{\rho }^2\ge 0$. Initially $N^{\prime }(0)=2\alpha N_0>0$, so $N$ stays positive and its logarithm is defined. The Cauchy-Schwarz inequality now givesTherefore $(\ln N{)}^{\prime }(t)\ge (\ln N{)}^{\prime }(0)=2\alpha$, and integration provesThis is an exponentially growing perturbation. Scaling the initial displacement and velocity down leaves $\alpha$ unchanged, so arbitrarily small initial perturbations display the same linear growth mechanism. It proves that a negative-energy direction gives exponential growth in a self-adjoint wave equation, without assuming a discrete spectrum or quoting a trial-function variational theorem.

Positive $W$ on nonzero displacements gives energetic stability; a negative direction gives instability. For mathematical precision, a uniform bound on the mass-weighted displacement norm follows if $W\ge \frac{1}{2}{\omega }_{\min }^2\Vert \mathbf{\xi }{\Vert }_{\rho }^2$ for some ${\omega }_{\min }>0$, giving $\Vert \mathbf{\xi }{\Vert }_{\rho }^2\le 2E/{\omega }_{\min }^2$. Strict positivity alone in an infinite-dimensional space need not supply such a uniform spectral gap; the standard magnetohydrodynamic energy principle asserts energetic stability and absence of growing modes. Cases with zero directions are marginal and require separate consideration.

The WKB approximation treats the vertical wavenumber as locally constant over one vertical wavelength. Write $w=\Re \{A(z)e^{ikx+i{\int }^{z}m(s)ds}\}$, with $|m^{\prime }|\ll m^2$ and $|A^{\prime }|\ll |mA|$. For the stationary internal-wave WKB solution, $A\propto m^{-1/2}$ away from singular or turning levels. A small terrain slope and small perturbations are also needed for Linearized Boussinesq equations.

Let $p$ be the perturbation fluid pressure divided by reference mass density, and $b$ the buoyancy perturbation. The steady Linearized Boussinesq equations and incompressibility condition areFor an $e^{ikx}$ component, elimination gives $u=-w^{\prime }/(ik)$, $b=-N^{2}w/(ikU)$ and $p=(U{w}^{\prime }-U^{\prime }w)/(ik)$. Substitution into vertical momentum gives the exact stationary Taylor–Goldstein equationThe leading WKB approximation therefore identifiesThe kinematic boundary condition fixes the horizontal wavenumber to $k=k_T$ for the sinusoidal forcing.

For explicit heights assume a constant positive buoyancy frequency $N$, $0<k{U}_0<N$, and $|S|/N\ll 1$. The latter is a large gradient Richardson number. The linear shear flow has $U^{\prime \prime }=0$, so the propagation condition is $|U|<N/k$.

For $S<0$, the first obstruction is the critical level of an internal gravity wave atAs $z\uparrow z_c$, $m\sim N/U\to \infty$. The critical-height approach of a stationary wave in linear shear has upward group velocity $c_{gz}\sim k{U}^2/N$, so the travel time to $z_c$ diverges: the height is a supremum, approached in the ideal ray model. Here $|m^{\prime }|/m^2\to |S|/N$, which can remain small; it would be incorrect to say that the short-wavelength WKB approximation necessarily fails solely because $m$ diverges. Nevertheless $w\propto U^{1/2}$ and displacement $w/(ikU)\propto U^{-1/2}$, so a finite disturbance eventually violates linearity. Dissipation and overturning also matter near this singular level.

For $S>0$, the turning height of a stationary wave in linear shear isAt $z_t$, $m=0$ and the WKB approximation fails. An Airy function matching region describes reflection and an upper evanescent wave. Thus $z_t$ is the highest height of vertical propagation, rather than a sharp cut-off of every disturbance.

The problem initially allows $N(z)$ to vary. Without the extra constant-$N$ assumption, replace the explicit turning height by the first root of $N(z)=k|U_0+Sz|$ encountered on a propagating branch. A declining $N(z)$ can cause a turning level before the negative-shear critical level. The two closed-form heights require the stated constant-stratification assumption.

Take a common constant buoyancy frequency $N$ and put $k=k_T>0$. Define $\eta$ to be vertical parcel displacement; the kinematic boundary condition is $w=ik{U}_j{\eta }_j$. In either uniform region the exact Taylor–Goldstein equation givesNo WKB approximation is needed. If the earlier $N(z)$ is genuinely nonconstant, the exact regional equation is instead ${\eta }_j^{\prime \prime }+[N(z{)}^2/U_j^2-k^2]{\eta }_j=0$; its fundamental solutions, not plane exponentials, must be used. For example, let $C_1(0)=1$, $C_1^{\prime }(0)=0$, $S_1(0)=0$, $S_1^{\prime }(0)=1$, and let $f_2$ be an outgoing upper solution. Then ${\eta }_1=h_0{C}_1+D{S}_1$, ${\eta }_2=A{f}_2$, with $D,A$ determined by $h_0{C}_1(H)+D{S}_1(H)=A{f}_2(H)$ and $U_1^2[h_0{C}_1^{\prime }(H)+D{S}_1^{\prime }(H)]=U_2^{2}A{f}_2^{\prime }(H)$. This gives the exact arbitrary-profile field implicitly, while a simple angle-only coefficient requires further stratification assumptions.

Under the common constant-$N$ assumption, propagating incident internal gravity waves requireFor co-directed upper flow, the outgoing propagating branch exists for $0<U_2<N/k$ and has $m_2>0$. The radiation condition chooses $m_j>0$ because the laboratory vertical group velocity is positive on the negative intrinsic frequency branch. Define $\tan {\theta }_j=m_j/k$, so $\cos {\theta }_j=k{U}_j/N$.

At the velocity jump, the material-interface jump conditions for stratified inviscid shear flow require continuity of displacement and of fluid pressure. Vertical velocity itself is generally discontinuous: $w_j=ik{U}_j\eta$. In terms of displacement the pressure is $p_j=U_j^2{\eta }_j^{\prime }$. The displacement impedance for an internal wave at a velocity jump is $Z_j=U_j^2{m}_j$. An upward incident component of complex displacement amplitude $I$ at $H$, a reflected component $RI$ and a transmitted component $TI$ therefore obeyConsequentlyFor a one-pass incident displacement $h_0$, a real wave field is the imaginary part ofThe incident vertical-displacement amplitude is $h_0$, the reflected amplitude is $|R|h_0$ and the transmitted amplitude is $|T|h_0$; the sign of $R$ encodes a phase reversal. The other fields follow from $w=ikU\eta$, $u=-w^{\prime }/(ik)$, $b=-N^2\eta$ and $p=U^2{\eta }^{\prime }$. If a perfectly reflecting terrain boundary is enforced after the return wave arrives, replace the incident coefficient $h_0$ by $h_0/[1+R{e}^{2i{m}_{1}H}]$. The one-pass amplitudes use the permitted neglect of repeated terrain reflections.

For common $N$, $Z_j=(N^2/k)\cos {\theta }_j\sin {\theta }_j$. The physical vertical-displacement transmission coefficient is thereforeThe printed coefficient has a different denominator and cannot be obtained for this standard displacement convention. For example, when $U_2=U_1$, there is no interface scattering and the vertical amplitude stays $h_0$, whereas the printed expression gives $h_0/\cos {\theta }_1$. A total displacement magnitude would instead be the vertical amplitude divided by $\cos {\theta }_2$; with unequal velocities it still does not give the printed expression. The transmission formula needs an explicit correction or a different, presently unspecified amplitude/interface model.

For $U_2>N/k$, choose $m_2=i\kappa$ with $\kappa =(k^2-N^2/U_2^2{)}^{1/2}$: the transmitted response decays as $e^{-\kappa (z-H)}$ and the same impedance equations give a complex $R,T$, with $|R|=1$. At $U_2=0$ the steady equation is singular. Oppositely directed upper flow formally supports propagating waves when $0<|U_2|<N/k$, but upward radiation then requires $m_2<0$ and signed $Z_2$; the positive-angle co-directed formula is not valid there. A real velocity jump can also have Kelvin-Helmholtz instability, which is excluded from this formal steady scattering calculation.

Let the free surface be at $r=R_1$ and take the Newtonian gravitational potential to vanish at infinity. Mass conservation fixes $R_1=(3M_1/(4\pi {\rho }_1){)}^{1/3}$. The enclosed mass is $m(r)=4\pi {\rho }_1{r}^3/3$, and hydrostatic equilibrium requires $d{p}_0/dr=-{\rho }_{1}Gm(r)/r^2$. With vacuum outside, $p_0(R_1)=0$, givingThe pressure is zero outside. A prescribed constant external pressure would simply add that constant to $p_0$.

The bulk Eulerian and Lagrangian fluid perturbations obey $\delta \rho =0$ because the mass density is uniform and the Lagrangian displacement has zero divergence. The linearized Euler momentum equation in the interior isTaking the curl gives ${\omega }^2\nabla \times \mathbf{\xi }=0$. Thus every nonzero-frequency normal mode has an irrotational flow displacement; the simply connected space formed by the interior admits $\mathbf{\xi }=\nabla U$ and incompressibility gives the Laplace equation ${\nabla }^{2}U=0$. For a regular center, one spherical harmonic component is $U=A_l{r}^l{Y}_l^m$, with $|m|\le l$.

It is essential to retain the surface gravity perturbation of a uniform-density sphere. Although the bulk mass density perturbation is zero, the displaced density discontinuity givesThe Poisson equation implies that $\delta \Phi$ is continuous and its outward radial derivative has the jumpUsing regularity at the center and decay at infinity, writeThe derivative jump is $-(2l+1)C_l{R}_1^{l-1}{Y}_l^m$, so $C_l=-4\pi G{\rho }_{1}l{A}_l/(2l+1)$.

Integrating the bulk Euler momentum equation gives $\delta p/{\rho }_1={\omega }^{2}U-\delta \Phi$ for $l\ge 1$. At the free surface the Lagrangian pressure perturbation vanishes:Combining these relations yields the Kelvin stellar mode frequencyThe $2l+1$ choices of $m$ are degenerate because of spherical symmetry. For $l=1$, $r{Y}_1^m$ is a linear combination of coordinates in a Cartesian coordinate system, so $\nabla U$ is a constant displacement: the zero frequency is rigid translation of the whole isolated body. Such a translation has no restoring force. The constant $l=0$ potential generates no displacement; a radial breathing motion is excluded by incompressibility and a regular center.

The printed irrotational assertion needs the nonzero-frequency qualification, or restriction to the potential incompressible stellar surface modes. It is false for all neutral displacements: $\mathbf{\xi }=\mathbf{\Omega }\times \mathbf{r}$ has zero divergence, zero normal displacement at the surface, and $\delta p=\delta \Phi =0$, so it is a zero-frequency displacement, but $\nabla \times \mathbf{\xi }=2\mathbf{\Omega }\ne 0$. This does not change the Kelvin stellar mode spectrum just derived.

The separation of the two bodies is $a$, as is fixed by the supplied Kepler third law ${\omega }_o^2=G(M_1+M_2)/a^3$. The phrase “orbit of radius $a$” cannot denote the first body's distance from the center of mass in that formula.

Expand the companion's Newtonian potential of a point mass about the first body's center. The constant term produces no force; the linear term produces the uniform acceleration of the center and disappears in the translating frame. The leading differential acceleration therefore comes from the quadrupolar tidal forcingwhere $\mathbf{n}(t)$ points toward the companion. It is a degree-two solid spherical harmonic. For a circular nonrotating-frame orbit, its time-dependent components have $m=\pm 2$ and frequency $2{\omega }_o$, because the squared direction cosines repeat twice per orbit. There is also a static $m=0$ component. The real part of one complex harmonic represents the pair of time-dependent components; its normalization and orientation are absorbed into the order-one coefficient $\psi$. This gives $\Psi =\psi (G{M}_2/a^3)r^2{Y}_2^m$ and $\omega =2{\omega }_o$. The expansion is the leading term for $R_1/a$ small; spin is being neglected.

For a forced potential flow with the same temporal frequency, the Euler momentum equation becomesUse $U=A{r}^2{Y}_2^m$. The surface gravity perturbation of a uniform-density sphere is unchanged, as is the vanishing Lagrangian pressure perturbation boundary condition. Integrating the equation and applying that condition givesSince ${\xi }_r(R_1)=2A{R}_1{Y}_2^m$, the real surface displacement isIndeed $ϵ{\omega }_2^2=2G{M}_2/a^3$, so the normalization is consistent. This is a particular forced response away from the resonance; arbitrary free normal modes can be added.

Put $\mu =M_2/M_1$ and $q=(R_1/a{)}^3$. The tidal resonance of a stellar oscillation condition is $4{\omega }_o^2={\omega }_2^2$, henceThe resonance gives $0\le ϵ<1/2$ for finite nonnegative $\mu$, with actual forcing requiring $M_2>0$. At exact resonance the undamped linearization has no bounded response at the forcing frequency: the time-domain particular solution grows proportionally to $t$ times a sinusoid. Thus the frequency-response formula has a pole, rather than describing a finite resonant displacement. Away from that pole, linearization requires the displayed surface displacement to be small compared with $R_1$; the leading quadrupolar tidal forcing approximation separately requires $R_1/a$ small.

Evaluating the function at the endpoint gives the displacementThereforeThe force-induced displacement $h$ is a linear map of $F$.

Taking the derivative of the force-dependent energy givesThus the derivative of the additional minimized internal energy with respect to the applied force equals the resulting endpoint displacement. This is the endpoint force derivative of clamped-free beam energy.

Subtracting the first row from the other two in the determinant givesThis vanishes exactly when the two displacements $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$ are linearly dependent. If $\mathbf{a}\ne \mathbf{b}$, that means $\mathbf{c}-\mathbf{a}=t(\mathbf{b}-\mathbf{a})$ for some real number $t$. Thus all three points lie on the lineConversely, putting three points on any one line makes their displacements linearly dependent, so the determinant vanishes.

The equivalence remains true without distinctness. If exactly two points coincide, the two distinct locations determine a line containing all three, and two rows of the original determinant coincide. If all three points coincide, any line through that point works and the determinant is again zero. Existence persists, although uniqueness of the line can fail.