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Let $A_0=f_0(0)$, $\delta =v_2-v_1$, $V=v_0+\delta$ and $K=k^2{\lambda }_{\mathrm{De}}^2$. Use the Debye length convention of the paper, ${\lambda }_{\mathrm{De}}=v_0/{\omega }_{\mathrm{pe}}$. On the candidate intervals in a, the Landau contour contributes no imaginary part. The piecewise constant slopes giveSince $n_0=v_0{A}_0$, setting the dielectric function to zero yieldsThe absolute values express the real response on the allowed intervals. Solving this real equation inside a sloping portion of $f_0$ would not satisfy the imaginary part of the dielectric function.

An exact algebraic form will also make the number of undamped branches clear. DefineOn each allowed interval, the quantity inside the logarithm equals $u^2(u-v_2)/[(u+v_0)(u-v_1)(u-V)]$. Exponentiating and usinggives the cubic equationFor every $k>0$, the logarithm runs from $0$ to $+\infty$ on $(-\infty ,-v_0)$, from $+\infty$ to $-\infty$ on $(v_1,v_2)$, and from $+\infty$ to $0$ on $(V,\infty )$. The intermediate value theorem supplies a root in each interval, and the cubic equation permits at most three. Thus there is exactly one undamped branch in each interval.

For the two outer branches, $|u|\gg v_0$ at small $k$. An integration by parts and a large-$u$ expansion giveThere is no boundary contribution because the phase-space distribution function vanishes at both endpoints. Direct area integration givesThus the exact small-$k$ cutoff, retaining the paper's definition $n_0=v_0{A}_0$, is $\omega \to \pm {\omega }_{\mathrm{pe}}\sqrt{n_{\ast }/n_0}$. To leading order in the narrow-velocity plateau approximation this isThese are Langmuir waves. The difference between $n_{\ast }$ and the reference $n_0$ is only first order in $\delta /v_0$, so it does not alter the requested leading-order result.

For the finite-phase velocity branch put $u=v_1+\delta a$, with $0<a<1$. To leading order in $\delta /v_0$ and $\delta /v_1$, the real dispersion relation becomesAt long wavelength set $K=0$ to leading order. Then $a=v_1^2/v_0^2$, andFor comparison, the exact $k\to 0$ phase velocity from the cubic equation in b is $B/A$; expanding it gives precisely $v_1+\delta v_1^2/v_0^2+O({\delta }^2/v_0)$ at fixed $v_1/v_0$.

A single Maxwell-Boltzmann velocity distribution has a nonzero slope at every finite nonzero $u$ and has no velocity plateau. The would-be electron acoustic wave is therefore subject to strong Landau damping, so there is no corresponding completely undamped acoustic branch in a single Maxwellian electron population with immobile ions. At its only stationary point $u=0$, a Maxwellian has $\int f_0^{\prime }(v)/vdv<0$, making the static dielectric function positive rather than zero. This statement does not exclude acoustic modes of suitably distinct electron populations or an ion-supported mode when ion dynamics is retained.

For $k>0$, the upper Langmuir wave starts at $\omega \approx +{\omega }_{\mathrm{pe}}$ and approaches the line $\omega =kV$ from above. The lower Langmuir wave starts at $\omega \approx -{\omega }_{\mathrm{pe}}$ and approaches $\omega =-k{v}_0$ from below. The electron acoustic wave starts at the origin with slope $v_1+\delta v_1^2/v_0^2$ and approaches $\omega =k{v}_1$ from above; it stays inside $k{v}_1<\omega <k{v}_2$. These are the only completely undamped branches, as shown by the cubic equation in b.

The small asymmetry between the outer branches reflects the extra positive-velocity population introduced by the velocity plateau. The negative-$k$ sketch follows by rotating the displayed branches through the origin, since $ϵ(-\omega ,-k)=ϵ(\omega ,k)$ on these undamped solutions. One should not impose symmetry about the $\omega =0$ axis at fixed positive $k$.

For the neo-Hookean solid, the axial Cauchy stress tensor under the unloaded lateral boundary conditions isWith $K,\mu >0$, the spring balance is equivalent, after multiplication by $\alpha >0$, to the cubic equationFor a fixed $\gamma \ge 1$, write its left side as $h(\alpha )$. It obeys $h(0)=-\mu <0$ and tends to $+\infty$ as $\alpha \to \infty$. MoreoverThus $h$ first decreases from a negative value and then increases strictly, crossing zero exactly once on $\alpha >0$. For $\gamma >1$,so the unique equilibrium has $1/\gamma <\alpha <1$. The tissue is elastically compressed, while its actual length $\alpha \gamma$ exceeds one and compresses the spring.

To determine the large-growth limit without presuming a power law, solve for $\gamma$:This is a strictly decreasing function of $\alpha >0$, since its derivative isIt tends to $+\infty$ only as $\alpha \to 0^{+}$. Consequently unlimited growth forces $\alpha \to 0^{+}$, andMore precisely, the dominant balance is $K\gamma {\alpha }^3\sim \mu$, givingThe surrounding spring softens the divergence relative to rigid confinement but does not bound the compressive stress. Both axial extension and lateral expansion help accommodate growth:whose product is the total volume $\gamma$, as required by elastic incompressibility and the imposed growth tensor.

The shell equation is conservative because its radial acceleration is the negative derivative of the specific potential energyMultiplying the equation of motion by $\dot{r}$ givesHenceThis is energy per unit shell mass. The total energy of a uniform sphere requires integrating the energies of its shells.

For a sphere of physical radius $R$, the shell theorem gives $M(r)=M(r/R{)}^3$ and $dM=3M{r}^{2}dr/R^3$. Assemble the sphere shell by shell to obtain its Newtonian gravitational potential energy:The cosmological constant contributes a prescribed one-body potential $-\Lambda r^2/6$, rather than a pair interaction, so there is no additional factor $1/2$:Evaluating at the turnaround radius givesAs in the question, $\Lambda$ here includes the factor $c^2$ that would multiply the geometrical cosmological constant in SI units.

For the final equilibrium apply the scalar virial theorem. The gravitational virial is $W_G$, whereas the cosmological constant force contributesThus, neglecting a surface-pressure term,At fixed $R$, a positive cosmological constant reduces the kinetic energy needed for virial equilibrium because its outward acceleration partially offsets self-gravity.

For the uniform-sphere collapse with a cosmological constant, assume a uniform final sphere of the same mass, with no energy loss or escaping matter. PutThe sphere has no bulk kinetic energy at turnaround, so $E_{\mathrm{ta}}=-A-B$. The final virial theorem givesSince ${\rho }_{\mathrm{ta}}=3M/(4\pi r_{\mathrm{ta}}^3)$,Conservation of energy therefore givesSelect the collapsing branch continuous from $x=1/2$ at $\eta =0$, rather than an unrelated root of the cubic equation.

For $|\eta |\ll 1$, the root lies close to $1/2$. Linearize $x^3$ there:Substitution into the cubic equation gives $(\eta /2-2)x+1-\eta /2\simeq 0$, whenceThe rational expression is an approximation, not an exact solution of the cubic equation. In fact both it and the exact collapsing root agree through order ${\eta }^2$; their first difference is of order ${\eta }^3$.

With zero cosmological constant, spherical collapse gives $r_f=r_{\mathrm{ta}}/2$. For small positive $\Lambda$, the same sphere with the same turnaround radius reaches a smaller final radius. This does not mean that repulsion promotes collapse: as the sphere contracts, $W_{\Lambda }$ becomes less negative, so some released gravitational binding energy pays the energetic cost of moving inward against the repulsive force. Combined with the modified virial theorem, energy conservation requires a more compact endpoint. Positive $\Lambda$ can prevent turnaround altogether; an inward acceleration at $r_{\mathrm{ta}}$ requiresThe perturbative radius formula applies well inside this limit, for spheres that actually turn around. A small negative cosmological constant instead gives $r_f/r_{\mathrm{ta}}>1/2$ at fixed turnaround data.

Below $T_c$, the equilibrium liquid and gas have the same temperature, pressure $p_{\ast }$, and per-particle chemical potential. For macroscopic bulk thermodynamic phases, neglect interface contributions. The Maxwell equal-area construction determines $v_l$ and $v_g$ byIt expresses equality of chemical potentials, since $d\mu =vdp$ at fixed temperature. Equivalently, the equilibrium Helmholtz free energy per particle follows a common tangent line between the two endpoint states, lying below the homogeneous continuation between them. The construction uses that mathematical continuation to compare endpoints; it is not a physical compression path through negative isothermal compressibility states.^[ref]

Here is a reproducible calculated example at $t=0.9$. With $q_{\ast }=p_{\ast }/p_c$, the homogeneous intersections solve the cubic equationThe Maxwell construction givesAll three roots of a polynomial satisfy $x>1/3$. Only the smallest and largest are the phase coexistence endpoints; the middle one has negative isothermal compressibility. Equal pressures at three admissible values of volume therefore do not imply three stable thermodynamic phases. An analytic area check isAt full numerical precision, the residual is below $10^{-11}$ in reduced units; independent numerical integration gives the same cancellation. At $t=0.9$, the spinodal points are instead $x\simeq 0.718597189$ and $1.528504964$.

The equilibrium plateau changes the phase fraction rather than the endpoint densities. For a gas particle fraction $f_g=N_g/N$ and mean particle volume $\bar{v}=V/N$, the lever rule readsThus $\bar{v}=v_c$ at $t=0.9$ gives $f_g\simeq 0.227219$: this is a particle fraction, not the volume fraction occupied by gas. Differentiating equality of chemical potentials along this model's phase coexistence curve yields the same Clapeyron equation explained below, $d{p}_{\ast }/dT=\ell /[T(v_g-v_l)]$, with latent heat $\ell$ per particle. Both endpoint differences vanish as the Van der Waals critical point is approached; the equation's critical limit must be taken as a ratio.

The Maxwell construction presumes a pure bulk fluid at thermodynamic equilibrium. Its endpoint equality does not specify how rapidly a real compression reaches equilibrium. A homogeneous metastable sample can persist while nucleation is delayed; the spinodal marks loss of local mechanical stability in this mean-field theory, rather than the equilibrium onset of condensation. Surface tension, finite sample size and impurities can affect the observed transition. These qualifications matter when comparing a mathematical Van der Waals isotherm with a visible liquid meniscus.

Consider only neutral $\mathrm{HA}$ added to pure water, at fixed temperature, with ideal dilute thermodynamic activities, water thermodynamic activity approximated by one, and no added salts or other chemical species. Normalize every concentration by $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$:Conservation of acid units and zero net electric charge, respectively, requireThe second equation balances positive hydronium against negative conjugate base and hydroxide. It would need all additional ions if salts or a chemical base were added. Together with acid dissociation constant $K_a=hx/y$ and the self-ionization of water relation $K_w=ho$, these giveThis is exact within the stated ideal model, including depletion and self-ionization of water. Multiplication by $h(K_a+h)$ yields the cubic equationFor positive $K_a,K_w,c$, the unmultiplied left-hand side increases strictly from negative to positive values as $h$ increases from zero to infinity; its derivative is $1+K_w/h^2+K_{a}c/(K_a+h{)}^2>0$. There is therefore exactly one positive root of a polynomial. A bisection method can find it without choosing an unphysical root of a polynomial.

If $K_w/h^2\ll 1$, neglect hydroxide, so $h\approx x$ and the quadratic equation retaining acid depletion givesThe last form avoids subtracting nearly equal numbers. If also $x/c\ll 1$, neglect depletion to obtainCheck both conditions after computing: $x/c$ measures the fraction lost from the initial undissociated weak acid inventory, while $K_w/h^2$ measures the omitted hydroxide contribution to electric charge balance. A $5\%$ depletion criterion is a chosen tolerance, not a universal boundary. Keeping depletion does not cure the neglect of self-ionization of water. These weak acid mass and charge balances also explain why the square root estimate cannot be extended to arbitrarily dilute chemical solutions.