Average the trial field over an integer number of periods. The squared-field, squared-gradient and squared-Laplacian averages are $A^2/2$, $q_0^2{A}^2/2$ and $q_0^4{A}^2/2$, while $⟨{\varphi }^2|\nabla \varphi {|}^2⟩=A^4{q}_0^2/8$. Hence the single-mode smectic mean-field free energy isFor the physically stabilizing case $B>0$, minimization givesThe amplitude tends continuously to zero at $a_c$, so this mean-field approximation predicts a continuous phase transition. The ordered minimum has free-energy density $-(a_c-a{)}^2/(8B{q}_0^2)$.

A nonzero $B$ alone is not sufficient for this conclusion: $B<0$ makes the displayed quartic amplitude energy unbounded below and requires higher-order stabilization. The continuous-transition statement uses $B>0$ and the stipulated fixed wavevector $q_0$.

Take $\gamma >0$ so that the short-wavelength quadratic energy is stable. For $B=0$, a Fourier mode with wavevector magnitude $q$ has inverse Gaussian field theory covarianceFor $\kappa <0$, its minimum is atThe zero-average constraint excludes the uniform mode but allows $\varphi =A\cos (q_{0}z)$. At $a=a_c$ the modes on the nonzero-wavevector soft-mode sphere become soft; below it they have negative quadratic energy, so the homogeneous state loses stability. A modulation along any chosen direction is the simplest smectic phase pattern. The quadratic instability selects its wavelength; with $B=0$, a nonlinear stabilizing interaction would still be needed to determine its amplitude.

The mass matrix is a positive-definite matrix, and integration by parts with periodic boundary conditions gives $C^T=-C$. HenceThis is energy conservation for semidiscrete Galerkin advection: $U^{T}MU=\Vert u_h{\Vert }_{L^2(0,1)}^2$. Thus the semidiscretization is stable and conserves its finite element L2 norm.

For a mesh-independent comparison with the grid norm, the Fourier symbol of $M$ is $h(2+\cos \theta )/3$, between $h/3$ and $h$. ConsequentlyThe conserved energy therefore gives a uniform bound in $\sqrt{h{\sum }_m|U_m{|}^2}$ as well. Equivalently, each Fourier mode evolves with the purely imaginary exponent $\sigma (\theta )=3i\sin \theta /[h(2+\cos \theta )]$. This concerns the semidiscrete method; a chosen time integrator must separately control those imaginary eigenvalues.

Let $x_m=mh$ be a uniform periodic mesh, with indices interpreted modulo the number of nodes. Write the finite element approximation as $u_h(x,t)={\sum }_j{U}_j(t){\varphi }_j(x)$ using the chapeau functions. The Galerkin method requiresThe mass matrix has $M_{mm}=2h/3$ and $M_{m,m\pm 1}=h/6$. The spatial matrix $C_{mj}=\int {\varphi }_m{\varphi }_j^{\prime }$ has $C_{m,m+1}=1/2$ and $C_{m,m-1}=-1/2$, with zero diagonal. Therefore the semidiscrete equations areor equivalently $M\dot{U}=CU$. The coefficients come from integrating the products of the two overlapping piecewise-linear hat functions on each interval; periodicity supplies the wraparound entries.

For a Fourier mode, the amplification polynomial isPut $a=1-2\mu$ and $\xi =e^{i\theta /2}\eta$. The roots becomeIf $|a|<1$, the square root is real, both roots have modulus one, and their separation is at least $2\sqrt{1-a^2}>0$. The companion matrix has eigenvectors $({\xi }_{\pm },1{)}^T$, so their uniform separation gives a mesh-independent bound on its powers. By power boundedness of a two-level Fourier scheme and the discrete Fourier transform, this proves stability for $0<\mu <1$.

If $|a|>1$, choose $\theta =\pi$. The roots have product of modulus one and unequal moduli, so one has modulus greater than one. The scheme is unstable.

The boundary cases require more than a root-modulus check. At $\theta =\pi$, $\mu =0$ gives $(\xi -1{)}^2=0$, while $\mu =1$ gives $(\xi +1{)}^2=0$. The companion matrices have nontrivial Jordan blocks and solutions of the form $(A+Bn)(\pm 1{)}^n$. Their linear growth rules out a uniform stability bound. This mode occurs on even periodic grids, which are enough to disprove mesh-uniform stability at the endpoints.

Therefore the stable range is exactlyWithin this range, second-order convergence additionally requires two appropriately accurate starting time levels. The unit-modulus parasitic root near $\theta =0$ does not decay, so poor initialization cannot be repaired by the scheme.

Keep the Courant number $\mu =\Delta t/\Delta x$ fixed and write $h=\Delta x$. An exact solution of $u_t=u_x$ is $u(x,t)=g(x+t)$. Substituting it into the update gives the defectwhere $s=x+t$ and $g^{(3)}$ is evaluated at $s$. The constant, first-derivative and second-derivative terms cancel. Dividing by $\Delta t=\mu h$ gives a second-order consistency error. Hence for fixed $0<\mu <1$, $\mu \ne 1/2$, the scheme has order two.

At the special value $\mu =1/2$, the update reduces to $u_m^{n+1}=u_{m+1}^{n-1}$. This is exactly the transport over two time steps, since $2\Delta t=\Delta x$, so the defect vanishes identically. With exact data at both starting levels it transports their values exactly; errors in those starting levels are still present, and the even and odd time subsequences evolve separately.

With $h=\Delta x$, $\tau =\Delta t$, $r=\tau /h^2$ and $c=\alpha \tau /h$, the Forward Euler method givesFor a Fourier mode $e^{im\theta }$, the amplification factor is $G(\theta )=1-4r{\sin }^2(\theta /2)+ic\sin \theta$. Set $s={\sin }^2(\theta /2)$. ThenThe expression in braces is an affine function of $s$, so it is nonpositive for all $s\in [0,1]$ precisely when its two endpoint values are nonpositive. For $\tau >0$, these conditions are $c^2\le 2r$ and $r\le 1/2$. Thus Forward Euler stability for centered advection-diffusion requires exactlyFor $\alpha =0$, only the first condition remains. If either bound fails, a band of Fourier modes has $|G|>1$, giving instability under von Neumann stability analysis. At equality the amplification bound still holds, so the endpoints are included. This is discrete L2 norm stability; nonnegative stencil weights would impose the additional spatial restriction $|\alpha |\Delta x\le 2$.

Write $h=\Delta x$ and set $u_0=u_{M+1}=0$. In the discrete L2 norm $\Vert u{\Vert }_h^2=h{\sum }_{m=1}^M|u_m{|}^2$, the energy method givesTo obtain this identity, shift the indices in the centered second-difference sum. The centered first-difference matrix is skew-symmetric, so its contribution has zero real part even for complex grid values. The homogeneous Dirichlet boundary conditions remove the endpoint terms.

Therefore the semidiscretization is stable for every real $\alpha$ and every positive $h$, with the mesh-independent bound $\Vert u(t){\Vert }_h\le \Vert u(0){\Vert }_h$. No restriction on $|\alpha |h$ is needed for this energy estimate.

For algebraic stability, the weights must be nonnegative and the real symmetric matrix $M=BA+A^{T}B-b{b}^T$, where $B=\mathrm{diag}(b)$, must be a positive semidefinite matrix.

The weights are positive, but $a_{11}=0$ and $b_1=1/6$, soA positive semidefinite matrix cannot have a negative diagonal entry, as its quadratic form at the first standard basis vector would be negative. Thus this Lobatto IIIA method is not algebraically stable, despite its A-stability.

The stability function of a Runge-Kutta method is $R(z)=1+z{b}^T(I-zA{)}^{-1}1$. Substitution givesWriting the numerator and denominator as $N(z)$ and $D(z)$,The denominator has zeros $3\pm i\sqrt{3}$, both in the right half-plane. Therefore $|R(z)|\le 1$ throughout the closed left half-plane, with strict inequality in its interior. The method is A-stable. Since $R(z)\to 1$ at infinity, it is not L-stable.

The tableau is the three-stage Lobatto IIIA method. For its stage matrix $A$, weights $b$, nodes $c$ and $C=\mathrm{diag}(c)$, the Butcher order conditions through order four arewhere $c^{\circ j}$ means coordinatewise powers. Direct substitution verifies all eight identities. Equivalently, this is a collocation Runge-Kutta method at the three Lobatto quadrature nodes, whose weights integrate cubic polynomials exactly.

Order five would require $b^T{c}^{\circ 4}=1/5$, but here $b^T{c}^{\circ 4}=5/24$. Therefore the method has exactly order four.

For the Dahlquist test equation, put $z=h\lambda$. The amplification roots satisfyWe show that no root can reach the unit circle for $\mathrm{Re}z<0$. If $\xi =e^{i\theta }$, solving the quadratic in $z$ givesWrite $c=\cos \theta$, $u=\mathrm{Re}C=-3/4+4c-c^2$, and $v=\mathrm{Im}C=(c-4)\sin \theta$. Then $u\le 9/4$ andSince $9/2-u>0$, this gives $|C|+u\le 9/2$. Consequently $|\mathrm{Re}\sqrt{C}{|}^2=(|C|+u)/2\le 9/4$, so both possible $z$ have nonnegative real part. Equality can occur only at $\theta =0$, which gives $z=0$ or $z=3$.

The leading coefficient has zeros $(3\pm i\sqrt{5})/2$, both in the right half-plane. Hence the amplification roots vary continuously as a pair throughout the left half-plane. At $z=-1$ they are $1/3$ and $1/5$; neither can leave the unit disk without crossing the unit circle, which the preceding calculation excludes. The same conclusion extends to the imaginary axis, with strict inequality away from $z=0$. At zero the roots $1,1/7$ satisfy the simplicity requirement.

Thus the method is A-stable. Its third order does not contradict the usual second-order barrier, because it is a multiderivative multistep method.

Expand about $t=t_{n+2}$ and use $y^{\prime \prime }=f^{\prime }(y)f(y)$ from the chain rule. The residual of the exact solution isThe coefficients through $h^3$ vanish, and the displayed fourth-order coefficient does not. Thus the multiderivative multistep method has order three.

At $h=0$, its first characteristic polynomial isIts roots are $1$ and $1/7$, with the unit-modulus root simple, so it satisfies the root condition for a multistep method and is zero-stable. Combining this with the defect estimate proves convergence of order three, assuming sufficiently smooth $f$, order-three starting values, and the nearby branch of each implicit update.

More explicitly, the right-hand side is $h{F}_h(y_{n+2})$, where $F_h=6f/7-2h{f}^{\prime }f/7$ has a uniform local Lipschitz continuity bound for small $h$. A bounded zero-stable impulse response and the discrete Gronwall inequality therefore control the accumulated $O(h^4)$ defects by $O(h^3)$ over fixed time intervals. This is convergence of a zero-stable multiderivative method, rather than a direct application of the first-derivative-only Dahlquist equivalence theorem.

Let $w=(1,2,1,0,0{)}^T$. Direct multiplication gives $Hw=(0,0,0,2,2{)}^T$, so $w^{T}Hw=0$.

Suppose $H=P+N$, where $P$ is a positive semidefinite matrix and $N$ is a symmetric nonnegative matrix. Since $w\ge 0$, both $w^{T}Pw$ and $w^{T}Nw$ are nonnegative. Their sum is zero, so both vanish. In particular,Every coefficient $w_i{w}_j$ in this sum is strictly positive and every $N_{ij}$ is nonnegative, forcing $N_{ij}=0$ for $1\le i,j\le 3$.

Now apply the same argument to all five cyclic shifts of $w$. The Horn copositive matrix is cyclically invariant, so every shifted vector also has zero quadratic form. It follows that the entries of $N$ vanish on every cyclic block of three consecutive indices. Every pair of indices on a five-cycle lies in such a block, hence $N=0$.

This would imply $P=H$. But the zero quadratic form of a positive semidefinite matrix would then force $Hw=0$, contradicting $Hw=(0,0,0,2,2{)}^T$. Therefore $H$ lies outside the positive-semidefinite-plus-nonnegative cone. By the previous equivalence, its quartic form is a nonnegative polynomial that is not a sum of squares polynomial.

The Horn copositive matrix is invariant under cyclic permutation of its five coordinates: its negative entries correspond precisely to neighboring indices on the five-cycle. For any $x\ge 0$, cyclically relabel the coordinates so that $x_5$ is a smallest coordinate. In particular, $x_4-x_5\ge 0$.

Using the stated identity in this coordinate order,The square is nonnegative, and both remaining terms are nonnegative since $x_i\ge 0$ and $x_4\ge x_5$. Cyclic invariance means the relabeling has not changed the quadratic form. Thus $H$ is a copositive matrix for every original ordering of $x$.

First suppose $A=P+N$, with $P$ a real positive semidefinite matrix and $N$ a symmetric nonnegative matrix. Factor $P=B^{T}B$, and write $v(z)=(z_1^2,\dots ,z_n^2{)}^T$. ThenEach term is a square multiplied by a nonnegative scalar, so $p$ is a sum of squares polynomial.

Conversely, suppose $p={\sum }_{\ell }{q}_{\ell }^2$. Because $p$ is a degree-four homogeneous polynomial, the homogeneous sum of squares representation allows every $q_{\ell }$ to be quadratic and homogeneous. Explicitly, higher-degree parts cannot cancel in a sum of squares; constant parts vanish because $p(0)=0$, and the degree-two part ${\sum }_{\ell }(q_{\ell }^{(1)}{)}^2$ forces all linear parts to vanish. WriteSince $p$ is invariant under every coordinate sign change, sign averaging of a sum of squares over independent Rademacher random variables givesThe cross terms vanish because their sign products contain an odd power of at least one independent sign. SetThen $P$ is a positive semidefinite matrix and $N$ is a symmetric nonnegative matrix. Comparing the coefficients of $z_i^4$ and $z_i^2{z}_j^2$ gives $A=P+N$. This proves the sum of squares criterion for a biquadratic form.

For $z\in {\mathbb{R}}^n$, define $x_i=z_i^2$. Then $x$ lies in the nonnegative orthant andIf $A$ is a copositive matrix, this is nonnegative for every $z$, so $p$ is a globally nonnegative polynomial. Conversely, every $x\ge 0$ has the form $x_i=z_i^2$ by taking $z_i=\sqrt{x_i}$. If $p$ is globally nonnegative, then $x^{T}Ax=p(z)\ge 0$ for every such $x$, proving that $A$ is a copositive matrix. The map $z\mapsto (z_i^2{)}_i$ covers the entire nonnegative orthant, which is the key to both directions.

For a proper three-graph colouring $c$, associate the colors with the three roots of unity $d(i)=\exp (2\pi i(c(i)-1)/3)$, viewed as unit vectors in ${\mathbb{R}}^2$. Define $U_{ij}=2⟨d(i),d(j)⟩$ using the real inner product.

This is twice a Gram matrix, so it is a positive semidefinite matrix. Its diagonal is $2$. On every edge, the endpoint colors differ, so their vectors make angle $2\pi /3$ or $4\pi /3$ and $U_{ij}=2\cos (2\pi /3)=-1$. Thus $t=3$ and this $U$ are feasible in the second semidefinite program, provingThis is the regular simplex construction of a strict vector coloring.