= Solution
Take a long periodic section of length $L$, or neglect end effects and choose an integer number of wavelengths. The amplitude $u_q$ here is a single real sine amplitude, fixing the normalization of the <thermal membrane undulation> spectrum. For a <lipid vesicle> with an <axisymmetric membrane deformation>, conservation of the enclosed volume gives
$$
\pi\int_0^L R(z)^2\,dz=\pi L\left(\rho_0^2+\frac{u_q^2}{2}\right)=\pi LR_0^2,
\qquad
\rho_0=\sqrt{R_0^2-u_q^2/2}=R_0-\frac{u_q^2}{4R_0}+O(u_q^4).
$$
The mean radius therefore decreases at second order. This adjustment is essential: fixing the mean radius instead of the volume would miss the unstable term in the <fixed-volume capillary spectrum of a cylindrical membrane>.
The <surface area> of an axisymmetric graph is $\mathcal S=2\pi\int R\sqrt{1+R_z^2}\,dz$. Expanding for $|u_q|\ll R_0$ and $q|u_q|\ll1$ gives
$$
\mathcal S=2\pi L\rho_0+\frac{\pi LR_0}{2}q^2u_q^2+O(u_q^4)
=2\pi LR_0+\frac{\pi L}{2R_0}\bigl[(qR_0)^2-1\bigr]u_q^2+O(u_q^4).
$$
Multiplication by the <membrane tension> yields a quadratic <potential energy> $\Delta E=K_q u_q^2/2$, where
$$
K_q=\frac{\pi\sigma L}{R_0}\bigl[(qR_0)^2-1\bigr].
$$
For $qR_0>1$, the <equipartition theorem> gives the stable-mode <fixed-volume capillary spectrum of a cylindrical membrane>
$$
\boxed{\langle u_q^2\rangle=\frac{k_BT R_0}{\pi\sigma L\bigl[(qR_0)^2-1\bigr]}.}
$$
Here $k_B$ is the <Boltzmann constant> and $T$ is the <temperature>. The cosine amplitude has the same variance and is an independent real coordinate at quadratic order. If instead $R-\rho_0=\sum_{q\ne0}a_qe^{iqz}$ with $a_{-q}=a_q^*$, the corresponding positive-$q$ complex coefficient has $\langle|a_q|^2\rangle=k_BT R_0/[2\pi\sigma L((qR_0)^2-1)]$. This is the same <thermal membrane undulation> spectrum in a different <Fourier mode> normalization.
For $qR_0<1$, \b[the cylinder is unstable rather than having a negative fluctuation variance]. The quadratic <surface area> change is negative: a sufficiently long-wavelength modulation reduces area while retaining volume. This is the <Rayleigh–Plateau instability>, expressed here as pearling of a tension-dominated <lipid vesicle>. A cylinder supporting such modes has no unconstrained harmonic <thermal equilibrium> about the uniform state. The $q=0$ uniform radius change is forbidden by fixed volume, and $qR_0=1$ is marginal in the tension-only approximation. The finite length and endpoint constraints determine which nonzero <Fourier modes> are allowed.
The large <membrane tension> assumption controls modes with $qR_0$ of order one. For large $q$ the omitted <membrane bending modulus> contribution grows approximately as $\kappa q^4$ and overtakes the tension term when $\kappa q^2\gtrsim\sigma$. Thus the displayed tension-only spectrum requires $q\ll\sqrt{\sigma/\kappa}$ as well as small amplitudes and a positive stiffness. Close enough to the marginal wavelength, bending corrections must also be retained; their relative scale at $qR_0$ of order one is $\kappa/(\sigma R_0^2)$.
For the circular phase boundary in a <lipid bilayer>, the energy is its perimeter times the <line tension> $\gamma$. Let $\varphi$ be the polar angle and write $r(\varphi)=\rho_0+u_m\sin(m\varphi)$, where the periodicity requires an integer $m=qR_0\ge1$. Fixed enclosed area implies
$$
\frac12\int_0^{2\pi}r^2\,d\varphi=\pi\left(\rho_0^2+\frac{u_m^2}{2}\right)=\pi R_0^2,
\qquad \rho_0=R_0-\frac{u_m^2}{4R_0}+O(u_m^4).
$$
The perimeter, including the mean-radius change, is
$$
\mathcal P=\int_0^{2\pi}\sqrt{r^2+r_\varphi^2}\,d\varphi
=2\pi R_0+\frac{\pi}{2R_0}(m^2-1)u_m^2+O(u_m^4).
$$
Thus the <fixed-area capillary spectrum of a circular boundary>, again for a single real sine or cosine amplitude, is
$$
\boxed{\langle u_m^2\rangle=\frac{k_BT R_0}{\pi\gamma(m^2-1)}
=\frac{k_BT R_0}{\pi\gamma((qR_0)^2-1)},\qquad m\ge2.}
$$
The complex <Fourier mode> coefficient convention again divides this result by two. The mode $m=0$ is excluded by the fixed-area constraint, so the negative-stiffness interval $0<m<1$ is not an allowed mode of a closed circular boundary.
At $qR_0=1$, the zero stiffness is the <translation mode of a circular boundary>, not a shape instability. Displacing the centre by a small distance $h$ changes the radius to first order by $h\cos\varphi$ or $h\sin\varphi$ while leaving area and perimeter unchanged. For example, an exactly translated circle has
$$
r(\varphi)=h\cos\varphi+\sqrt{R_0^2-h^2\sin^2\varphi}
=R_0+h\cos\varphi-\frac{h^2}{4R_0}+\frac{h^2}{4R_0}\cos(2\varphi)+O(h^4).
$$
The higher harmonics complete a true <translation mode of a circular boundary>. \b[The $m=1$ denominator therefore represents free centre motion:] an unconfined domain's centre has no restoring <force> or finite equilibrium variance in an infinite membrane. Radius fluctuations about a recentered domain omit this mode; external confinement would give it a separate restoring stiffness.
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