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For spatially varying tension in filament bending, keep the derivative of the filament tension as well as the curvature term. The first variation isTherefore the Euler-Lagrange equation and fluctuation operator areFor real, sufficiently smooth $\sigma$, the boundary form is the bending boundary form minus $[\sigma (\overline{u}{v}^{\prime }-\overline{u^{\prime }}v){]}_0^L$. Because $\sigma$ vanishes at both ends, the four self-adjoint endpoint conditions for filament bending still apply. The natural endpoint force also reduces there to the bending shear term. Thus $K_{\sigma }$ is a self-adjoint fourth-order scalar differential operator on the same chosen domain, with compact resolvent.

Choose a real orthonormal basis ${\psi }_n$ of eigenfunctions, $K_{\sigma }{\psi }_n={\lambda }_n{\psi }_n$, and write $h={\sum }_n{b}_n{\psi }_n$. Using the endpoint conditions in integration by parts givesThe equipartition theorem now gives, on the strictly positive subspace,This is a formal modal construction; no explicit eigenfunctions are needed. Nonnegative filament tension makes the energy nonnegative. Any surviving zero-energy filament modes must again be fixed. If signed $\sigma$ permits compression, self-adjointness still holds but does not guarantee a canonical ensemble: sufficiently strong compression can create negative eigenvalues and Euler buckling of an elastic filament. For instance, on $A=L=1$, take $\sigma =-Cx(1-x)$ and the clamped trial function $h=x^2(1-x{)}^2$. Thenso the energy is negative when $C>252$, despite $\sigma (0)=\sigma (1)=0$. The equipartition theorem requires a stable positive quadratic energy, not merely a real modal spectrum.

Use the homogeneous equilibrium point condition $k({\rho }_0){\rho }_0=a_{0}f({\rho }_0)$. Evaluate $D_1,D_2$ at $(a_0,{\rho }_0)$ and abbreviateThe quantity $c$ is the chemical relaxation rate with the cell density held fixed. Derivatives of $D_1,D_2$ do not enter this linear stability analysis: they multiply spatial derivatives of the homogeneous background or products of perturbations. For a Fourier mode with time dependence $e^{st}$, put $z=|\mathbf{q}{|}^2$ and writeConsequentlyAt $z=0$, the eigenvalues are $0$ and $-c$. The neutral cell-density mode reflects mass conservation, not decay of every homogeneous perturbation.

Write $a=D_2>0$, $b=D_{\rho }>0$, $C=D_1{f}_0$, and interpret production physically as $f_0\ge 0$. If $c\ge 0$, the trace-determinant stability criterion shows that an unstable nonzero wavenumber exists precisely when $C>ac$. The unstable band isIf $c<0$, the chemical field is already unstable at zero wavenumber; arbitrarily small positive wavenumbers are unstable too. Since $C\ge 0$, this also implies $C>ac$. Thus, on the infinite plane, the undivided Keller--Segel aggregation threshold isFor $\kappa >0$, dividing by $D_2\kappa$ gives the requested formAt equality there is no strictly growing mode when $c\ge 0$; nonzero spatial modes decay. On a finite domain, a permitted nonzero wavenumber must actually lie in the unstable band. If one allows a signed production function, the displayed matrix and trace-determinant stability criterion remain valid, but the simplification using $C\ge 0$ must be revisited.

The printed hypotheses do not ensure $\kappa >0$. Positivity of the degradation rate $k$ alone is insufficient. For a concrete counterexample, take $k(\rho )=e^{-2\rho }$, $f(\rho )=e^{-2}$, $a_0={\rho }_0=1$, and $D_1=D_2=D_{\rho }=1$. The homogeneous equilibrium point condition holds, all rates and transport coefficients are positive, but $\kappa =c=-e^{-2}$. The printed left-hand side equals $-1$, although $\det M=z(z-2e^{-2})<0$ for $0<z<2e^{-2}$. These spatial modes grow. If $\kappa =0$, the printed expression is undefined. The undivided criterion and the matrix above resolve both cases.

To find the fastest-growing Keller--Segel mode, use the larger eigenvalueFor $C>0$ the square root is real. Put $p=a+b$ and $d=b-a$. Differentiation givesA positive maximizing wavenumber exists in either of two growing cases: $c\ge 0$, $C>ac$, or $c<0$, $C>-bc$. Equivalently, $C>\max (ac,-bc)$. In this range $C+dc>0$, andThus the stationary point is the unique maximum. The derivative condition yieldsSolving it, with the root that satisfies the unsquared derivative equation, givesFor $c=0$, the unequal-diffusivity expression simplifies to $z_{\ast }=C/[\sqrt{ab}(\sqrt{a}+\sqrt{b}{)}^2]$, also agreeing with the equal-diffusivity limit. The formula maximizes the full two-field growth rate; it makes no instantaneous-chemical approximation. Every direction of $\mathbf{q}$ with this length is equivalent by rotational symmetry.

If $c<0$ but $0\le C\le -bc$, the maximum instead occurs at $z=0$: the fastest mode is homogeneous, with $s_{\ast }=-c$ and infinite wavelength. For $C>0$, its initial derivative is $-b+C/(-c)\le 0$. If $C+dc\ge 0$ the derivative thereafter decreases, while if $C+dc<0$ it increases towards the still-negative large-$z$ limit; either way no positive-$z$ maximum is missed. At $C=0$, the two eigenvalues are simply $-az$ and $-c-bz$, giving the same conclusion. In a finite box, maximize $s_{+}$ over the allowed Fourier modes; excluding the homogeneous mode can change the selected length. In a stable parameter range there is no fastest-growing mode.

The first ratio compares the positive feedback loop “more cells produce more attractant, which draws in more cells” with spreading by cell diffusion and removal of chemical perturbations. Its numerator $D_1{f}_0$ measures chemotaxis together with attractant production; its denominator $D_2\kappa$ measures dispersal together with incremental degradation. This competition between directed chemotaxis and cell diffusion produces spatial aggregation. The second ratio compares the concentration dependence of chemical production, $a_0{f}_0^{\prime }$, with incremental degradation $\kappa$. Positive $f_0^{\prime }$ amplifies chemical fluctuations, whereas negative $f_0^{\prime }$ suppresses them. It changes the chemical relaxation available to the chemotaxis feedback loop and can itself destabilize the homogeneous chemical field. Chemical diffusion $D_{\rho }$ suppresses short scales and sets the selected wavelength, but does not change the infinite-plane long-wave threshold. These interpretations as ratios of stabilizing and destabilizing processes assume $\kappa >0$.

Write $a_k=g-\cos k$, $s_k=\sin k$ and $r_k=\sqrt{a_k^2+s_k^2}$. The Ising-chain Bogoliubov diagonalization uses the eigenvalues $\pm r_k$ of the two-by-two Nambu matrix. For distinct paired modes $k\ne -k$, choose $\cos {\theta }_k=a_k/r_k$, $\sin {\theta }_k=-s_k/r_k$ and ${\theta }_{-k}=-{\theta }_k$. The rotation $U_k=e^{-i{\theta }_k{\sigma }_y/2}$ satisfies $U_k^{†}{M}_k{U}_k=r_k{\sigma }_z$, and its first transformed component isThe opposite signs of the paired angles preserve the canonical anticommutation relations. Self-paired spinless fermion modes have zero pairing and are treated directly as occupied or empty number levels; they do not require this paired-angle formula.

Both $k$ and $-k$ occur in the sum, so the physical quasiparticle coefficient is $2J{r}_k$, not $J{r}_k$. ConsequentlyThis is the bulk quadratic result under the stipulated boundary simplification. Restoring the finite-chain parity sectors adjusts allowed modes and global excitation constraints. In the thermodynamic limit, the ground-state energy per site isFor $g\ge 0$, the bulk quasiparticle gap is $2J|g-1|$. The requested three regimes are shown in the original sketch and discussed separately below.

For a trajectory connecting the degenerate minima in infinite Euclidean time, the first integral gives $m{\dot{q}}^2/2=V(q)$. The quartic double-well instanton and its reverse areTheir action isThe instanton fluctuation prefactor $K$ has dimensions of inverse time. It incorporates nonzero Gaussian fluctuation eigenvalues and the translation zero-mode Jacobian; a conventional determinant expression iswith common endpoint regularization and the prime removing the translation mode. Only its positive quantum tunnelling rate $\kappa =K{e}^{-S_0/\hslash }$ is needed here.

In the dilute instanton gas, well-separated crossings alternate in direction. Integrating $n$ ordered centers gives ${\tau }^n/n!$. A path returning to the same well has $n$ an even number, and one connecting opposite wells has $n$ an odd number. The common leading harmonic endpoint factor follows from the supplied harmonic oscillator transition kernel:Expanding these functions gives the sums over even numbers and odd numbers, with the position-kernel normalization ${\mathcal{A}}_0$ that its abbreviated formula suppresses. Validity requires $S_0/\hslash \gg 1$, $\omega \tau \gg 1$ and $\kappa \ll \omega$, so typical crossing separation greatly exceeds an instanton width. Local anharmonic corrections replace the harmonic well energy by its perturbatively corrected value; the leading formula does not claim uniformly negligible relative error at arbitrarily large $\tau$.

The two exponents identify the double-well tunneling splitting:The superposition with an even spatial wavefunction is the lower state. Coherent quantum tunnelling removes the classical degeneracy, and the exponentially small splitting sets the long quantum tunnelling timescale.

An eigenvector of a matrix $A$ is a nonzero vector $v$ for which $Av=\lambda v$. The scalar $\lambda$ is its eigenvalue. For a fixed eigenvalue, the zero vector together with all corresponding eigenvectors forms the eigenspace $\ker (A-\lambda I)$. The requirement $v\ne 0$ prevents every scalar from being an eigenvalue merely because $A0=\lambda 0$.

Expanding the characteristic polynomial givesSolving the three nullspace systems gives the eigenvalues and eigenspacesFor example, at $\lambda =1$ the first two equations force $z=0$ and $x=y$; at $\lambda =2$ they force $x=0$ and $y=-z$; at $\lambda =-1$ they force $x=y=z$.

Putting these eigenvectors in columns proves their linear independence:Hence $A$ is a diagonalizable matrix over the reals.

If $Ax=b$ and $h\in \ker A$, symmetry gives $b\cdot h=(Ax)\cdot h=x\cdot Ah=0$, proving necessity. Conversely, expand $b$ in the orthonormal basis from part (i), with $A{y}_i={\lambda }_i{y}_i$. The zero-eigenvalue vectors span the kernel of a linear map. If $b$ is orthogonal to that kernel, then $b\cdot y_i=0$ whenever ${\lambda }_i=0$, so the remaining coordinates can be solved independently.

For a nonzero eigenvalue and any associated nonzero eigenvector $v$, symmetry gives $b\cdot v=\lambda x\cdot v$. Thus the component vector along $v$ isFor the unit vectors $y_i$, this coefficient is $(b\cdot y_i)/{\lambda }_i$. Consequently the general solution isThis proves sufficiency as well as the claimed kernel criterion. If the compatibility condition fails, there is no solution; if it holds, the freedom is exactly an arbitrary kernel vector.

View a possible complex eigenvector $v$ using the Hermitian inner product. For a real symmetric matrix, $A^{†}=A$, so $v^{†}Av$ is real. Since $v^{†}Av=\lambda v^{†}v$ and $v^{†}v>0$, every eigenvalue is real.

For eigenvectors $v,w$ with distinct eigenvalues $\lambda ,\mu$, symmetry givesso $v^{†}w=0$. In particular, real eigenvectors are orthogonal in the real dot product.

Use the permitted diagonalizability assumption to take bases of the real eigenspaces whose union spans ${\mathbb{R}}^n$. Apply the Gram-Schmidt process within each eigenspace. Its linear combinations remain eigenvectors of that same eigenvalue; vectors in different eigenspaces are already orthogonal. The resulting union is an orthonormal basis of eigenvectors, giving an orthogonal diagonalization of a real symmetric matrix.

The phase plane system is $\dot{x}=y$, $\dot{y}=-cy-\sin x$. Its equilibrium points are $(k\pi ,0)$ for every integer $k$. At such a point the Jacobian matrix and characteristic equation areFor even $k$, the equilibrium point is a stable focus for $0<c<2$, with eigenvalues $-c/2\pm i\sqrt{1-c^2/4}$; it is a stable node for $c>2$, with two distinct negative real eigenvalues. The excluded value $c=2$ gives the repeated critical case.

For odd $k$, the eigenvalues are $(-c\pm \sqrt{c^2+4})/2$. One is positive and one negative, so each is a saddle equilibrium for every $c>0$. These classifications follow from the linearization of a dynamical system, since all the relevant eigenvalues have nonzero real part for $c>0$. The even equilibria remain hyperbolic at $c=2$ too, although their linearization then has a repeated eigenvalue; that critical case is excluded from the requested classification.

For $c=1$, the even equilibria are stable foci with eigenvalues $(-1\pm i\sqrt{3})/2$, and the odd equilibria are saddle equilibria with eigenvalues $r_{\pm }=(-1\pm \sqrt{5})/2$. The local saddle directions are $y=r_{\pm }[x-(2j+1)\pi ]$.

The phase portrait shows clockwise spiraling into each even equilibrium: on its right-hand horizontal axis, the flow initially points down. The stable separatrices of the saddles divide attraction basins on the unwrapped angle axis, while the unstable branches flow into neighboring sinks. Initial conditions with enough mechanical energy can cross one or more potential crests before being captured. The curves repeat under $x\mapsto x+2\pi$. They are trajectories, not conservative energy contours: energy dissipation of a damped pendulum rules out nonconstant periodic orbits.