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First $N(A)$ is finite: by Cauchy-Schwarz, $\Vert Ax{\Vert }_2\le ({\sum }_{ij}{a}_{ij}^2{)}^{1/2}\Vert x{\Vert }_2$. It is nonnegative and absolutely homogeneous. If $N(A)=0$, every $A{e}_j=0$, so every column is zero and $A=0$; the converse is immediate. The triangle inequality for the Euclidean norm gives $N(A+B)\le N(A)+N(B)$ after taking suprema. Thus $N$ is a norm.

For any vector $x$, rescaling gives $\Vert Ax{\Vert }_2\le N(A)\Vert x{\Vert }_2$. Applying this twice givesFurthermore $|a_{ij}|\le \Vert A{e}_j{\Vert }_2\le N(A)$, proving the requested strict entry bound when $N(A)<\epsilon$.

For a polynomial in matrix entries, expand each monomial at $A+H$. The terms involving exactly one entry of $H$ give a linear map $Df(A)[H]$, with coefficients polynomial in the entries of $A$. Every remaining term has at least two entries of $H$ and is $O(N(H{)}^2)$ for $N(H)\le 1$, because $|h_{ij}|\le N(H)$. This proves differentiability. Its derivative is continuous as an operator-valued function: each coefficient is continuous, and for $N(H)\le 1$ every $h_{ij}$ is bounded by one, so the operator-norm difference is bounded by the sum of the coefficient differences. Coordinate continuity follows from the same entry bound. Hence every such polynomial function is continuously differentiable.

In the Leibniz expansion of $\det (I+H)$, the identity permutation supplies $1+{\sum }_i{h}_{ii}$ plus terms of degree at least two. Any nonidentity permutation moves at least two indices, so each of its terms has at least two $H$ entries. ConsequentlyIf $A$ is invertible, multiplicativity of the determinant and the preceding expansion giveThe adjugate identity gives $\mathrm{adj}A=\det (A)A^{-1}$. Thus $d^{\prime }(A)[H]=\mathrm{tr}((\mathrm{adj}A)H)$ for invertible $A$.

For singular $A$, take invertible $A_r\to A$, using the permitted density result. Both $d^{\prime }$ and the adjugate entries are continuous polynomial expressions, so taking limits for every fixed $H$ proves the derivative of the determinant formula on all matrices:Here the standard adjugate is the transpose of the cofactor array, as required by $(\mathrm{adj}A)A=(\det A)I$. In components this derivative is ${\sum }_{ij}{\mathrm{cof}}_{ij}(A)h_{ij}$, also valid at singular matrices.

The error $e(f)=f^{\prime \prime }(0)-f(-1)+2f(0)-f(1)$ is linear and annihilates all degree-at-most-two polynomials. The Peano kernel theorem with derivative order three givesFor $t\ne 0$, the second derivative of the truncated quadratic at $x=0$ is $1_{\{t<0\}}$. Its values at $x=-1,0,1$ then giveThe value at the single point $t=0$ is irrelevant to the integral. This is the centered second-derivative Peano kernel. Its absolute integral isConsequentlyIts changing sign explains why cubic polynomials also have zero error: their third derivative is constant and ${\int }_{-1}^{1}K=0$. No fourth derivative is needed for the claimed bound.

Use $\hat{f}(\lambda )=(2\pi {)}^{-1/2}{\int }_{\mathbb{R}}{e}^{-i\lambda x}f(x)dx$. The transform statement requires sufficient decay and regularity for the transform and its derivatives; functions in the Schwartz space suffice, and all the functions constructed below belong to that space. Integration by parts twice and differentiation under the integral giveTransforming the differential equation therefore givesThe differential equation by itself does not ensure that an ordinary Fourier transform exists: $f=e^{x^2/2}$ solves it with $\mu =1$ but grows too rapidly. Thus the usual transformability assumptions are implicit in this part.

Take fixed $E>0$ and define $k=\sqrt{2mE}/\hslash$. For sufficiently large $V_0$, the barrier region has $\kappa =\sqrt{2m(V_0-E)}/\hslash$ and solves ${\psi }^{\prime \prime }={\kappa }^2\psi$. Solving this constant-coefficient equation gives the exact transfer relationIn the specified limit, $a=U/V_0$ and $\kappa a\to 0$, whileThe transfer matrix therefore tends to $\left(\begin{array}{cc}1& 0\\ \eta & 1\end{array}\right)$. As the two boundaries collapse to the origin, this proves continuity and the derivative jump for a delta potential:The vanishing exterior phase shift $e^{ika}\to 1$ introduces no further factor.

Write the incoming and reflected waves as $e^{ikx}+r{e}^{-ikx}$ on the left and the transmitted wave as $t{e}^{ikx}$ on the right. Matching givesThe denominator is nonzero for real $\eta$ and $k>0$, so the limiting matching system is well defined. The probability currents on either side have the same factor $\hslash k/m$, making the transmission probability $|t{|}^2$. Thus scattering by a delta potential givesThe limit retains a finite reflection probability despite the barrier's vanishing width because its integrated strength remains finite.

Let ${\zeta }_n$ be a primitive nth root of unity. The cyclotomic polynomial isInduction on $n$ shows that ${\Phi }_n$ is monic with integer coefficients. Here is the essential irreducibility of cyclotomic polynomials argument. Let $f$ be the minimal polynomial of ${\zeta }_n$ and let $\ell \nmid n$ be a prime. If ${\zeta }_n^{\ell }$ had a different monic irreducible polynomial $g$, then $f\mid g(X^{\ell })$. Reducing modulo $\ell$ gives $\bar{f}\mid {\bar{g}}^{\ell }$, so $\bar{f}$ and $\bar{g}$ have a common factor. As $fg\mid {\Phi }_n$, this would give a repeated factor of $X^n-1$ modulo $\ell$, impossible because its derivative is $n{X}^{n-1}$. Thus ${\zeta }_n^{\ell }$ is conjugate to ${\zeta }_n$. Every positive integer prime to $n$ is a product of such primes, so all primitive nth roots of unity are conjugate. ConsequentlyIn particular the cyclotomic field is an abelian extension. Complex conjugation corresponds to $-1$; for $n>2$ its fixed field is $\mathbb{Q}({\zeta }_n+{\zeta }_n^{-1})$, the maximal real subfield, of degree $\phi (n)/2$.

The ring of integers is $\mathbb{Z}[{\zeta }_n]$. One way to see that there is no missing index is to work at each prime $p$. Write $n=p^{a}m$ with $p\nmid m$. The prime-to-$p$ roots of unity have separable reduction; their local factors generate unramified extensions with full integer rings. Over each such factor, adjoining ${\zeta }_{p^a}$ is generated by the Eisenstein polynomial ${\Phi }_{p^a}(1+T)$, with uniformizer ${\zeta }_{p^a}-1$, and hence gives the full integer ring of that totally ramified factor. The coprimality of $p^a$ and $m$ means that ${\zeta }_n$ generates both sets of roots of unity. Thus ${\mathbb{Z}}_p[{\zeta }_n]$ is maximal at every $p$, proving the integral assertion.

The same local description computes ramification and splitting. For $p\nmid n$, the ramification index is one, the residue degree is the multiplicative order of $p$ modulo $n$, and there are $\phi (n)/f$ primes above $p$. If $p\mid n$, put $m=n/p^a$; thenwhere ${\mathrm{ord}}_1(p)=1$. Indeed, the prime-to-$p$ part is unramified, and the $p$-power part is totally ramified by the preceding Eisenstein polynomial. Notice the exception $p=2,a=1$: its ramification index is one, consistently with $\mathbb{Q}({\zeta }_{2m})=\mathbb{Q}({\zeta }_m)$ for odd $m$. Thus finite ramification occurs only at the indicated conductor primes, with that exception.

We now prove cyclotomic Artin reciprocity over the rational numbers, using arithmetic Frobenius automorphisms. If $\ell \nmid n$ and $\mathfrak{L}$ lies over $\ell$, then Frobenius sends the reduction of ${\zeta }_n$ to its $\ell$th power. Distinct nth roots of unity remain distinct modulo $\mathfrak{L}$, since $X^n-1$ has separable reduction. Hence the Artin symbol is exactlyExtend multiplicatively to the group $I(n)$ of rational fractional ideals prime to $n$. Such an ideal has a unique positive rational generator $q$ whose numerator and denominator are prime to $n$. The formula isEvery unit residue modulo $n$ has a positive integer representative prime to $n$, so this Artin map is surjective. Its kernel is precisely the subgroup $P_{n\infty }$ generated by positive $q\equiv 1(\mathrm{mod}n)$. ThereforeThis proves the Artin reciprocity law for this extension, including its principal-ray kernel. The infinite component of the modulus records positivity and therefore the possible action of complex conjugation.

For a local field $F$ containing the Nth roots of unity, fix the Local Artin map with arithmetic Frobenius convention. The Hilbert norm residue symbol can be defined byIt is independent of the chosen root, continuous and bimultiplicative, factors through $F^{\times }/(F^{\times }{)}^N$ in each variable, and gives a nondegenerate pairing on that finite group. It satisfies $(a,b{)}_{F,N}=(b,a{)}_{F,N}^{-1}$ and $(a,1-a{)}_{F,N}=1$ for $a\ne 1$. Its norm interpretation isFor $N=2$ the quadratic Hilbert symbol is symmetric and sign-valued. It is also characterized by the existence of a nonzero solution of $z^2=a{x}^2+b{y}^2$. Moreover $(a,-a{)}_F=1$, and therefore $(a,a{)}_F=(a,-1{)}_F$. We now compute both the square-class groups and the pairing explicitly.

For odd $p$, write $a=p^{\alpha }u$ with $u\in {\mathbb{Z}}_p^{\times }$. An element is a square exactly when $\alpha$ is even and the residue of $u$ is a square in ${\mathbb{F}}_p^{\times }$. Necessity follows by reduction; sufficiency follows from the simple-root form of Hensel lemma applied to $X^2-u$, since $2x$ is then a unit. If $\eta$ is any unit with nonsquare residue, this givesThe nonsquare unit defines an unramified quadratic extension. Its field norms include every unit and have even valuation, as follows from the local norm calculation below. Thus $(\eta ,u{)}_p=1$ for every unit, but $(\eta ,p{)}_p=-1$. Finally $(p,p{)}_p=(p,-1{)}_p=(-1/p)$, by symmetry and the preceding unit calculation. Bimultiplicativity now proves explicit quadratic Hilbert symbols over the p-adic numbers:Here the factors are Legendre symbols of the nonzero residues. The formula works for arbitrary integer valuations $\alpha ,\beta$, since only their parity matters.

At $p=2$, the square of every odd integer is one modulo eight. Conversely, for an odd $u\equiv 1(\mathrm{mod}8)$, apply the strong form of Hensel lemma to $X^2-u$ at $1$: the error has valuation at least three, while the derivative has valuation one. Thus the unit square classes of the p-adic integers are precisely their four residues modulo eight. Combining this with parity of the valuation givesFor an odd unit putInspection of the residues $1,3,5,7$ gives $[u]=[-1{]}^{ϵ(u)}[5{]}^{\omega (u)}$. It remains to compute six basis pairings. Explicit field norms giveindeed $2=1^2+1^2$, $5=1^2+2^2$, $2=2^2-2\cdot 1^2$, and $5=5^2-5\cdot 2^2$.

In contrast $(-1,-1{)}_2=-1$. To justify the obstruction even for rational 2-adic coordinates, write $x=2^t{x}_0$, $y=2^t{y}_0$ with $x_0,y_0$ integral and at least one odd. Then $v_2(x_0^2+y_0^2)$ is zero if exactly one is odd, and one if both are odd. A sum of squares equal to the unit $-1$ consequently forces $t=0$ and exactly one coordinate odd. Its residue modulo four would then be one, whereas $-1$ is three.

Similarly $(2,5{)}_2=-1$. If $x^2-2y^2$ is a unit, the valuations of the two terms have opposite parity, so cannot cancel. It follows that $x$ is an odd integral element and $y$ is integral. Modulo eight its norm is then $1$ or $7$, never $5$. These signs, symmetry, and bimultiplicativity giveThe two boxed formulas determine the quadratic Hilbert symbol on all nonzero elements, including every rational prime $p$.

For unit resource requirements, treat the resource-blocking events as approximately independent. Let $B_j$ be the approximate blocking probability at resource $j$. Screening class-$r$ traffic by every resource other than $j$ produces the reduced-load approximationThe modeled resource is excluded from its own screening product. For general integer requirements, the standard generalized approximation is $a_j=(1-B_j{)}^{-1}{\sum }_r{A}_{jr}{\alpha }_r{\prod }_k(1-B_k{)}^{A_{kr}}$; in the unit-requirement case this reduces to the displayed formula. The approximate route acceptance is ${\prod }_j(1-B_j{)}^{A_{jr}}$. The independence used here is an approximation to the correlated exact stationary distribution, not an extra property of that law.

Here is a proof of fixed-routing existence and uniqueness. Suppose first that every retained capacity $C_j$ is a positive integer. Links of zero capacity permanently reject their routes and can be removed. If $K$ has the upper-truncated Poisson distribution with parameter $a$, its mean isThe derivative follows by differentiating the normalized weights: $dP(K=k)/da=P(K=k)(k-m_C(a))/a$. The probability at the upper endpoint similarly givesThus $E(\cdot ,C)$ increases from zero to one, while $m_C$ increases from zero to $C$. This is the monotonicity of the Erlang blocking probability.

Set $y_j=-\log (1-B_j)$. For $y\ge 0$ defineHere the inverse is with respect to offered load, and the value at zero is zero. This function is the carried load evaluated at the inverse blocking coordinate, so it is continuous and strictly increasing, with limit $C_j$ as $y\to \infty$. Consider the convex potential for the Erlang fixed pointEach exponential of a linear function is convex. Each integral is strictly convex because its derivative $u_j$ is strictly increasing. Their sum is strictly convex on the nonnegative orthant. It is also coercive: for large $y_j$, $u_j(y_j)$ is bounded below by a positive constant, so the integral grows without bound; the other terms are nonnegative. Hence $F$ attains a unique minimum.

At a positive coordinate, the first-order condition isAt $y_j=0$, the derivative equals minus the right-hand side and is nonpositive. A boundary minimum can therefore occur there only when that load is zero, in which case the derivative is zero and the same equation holds. Since $1-B_j=e^{-y_j}$, division by this positive acceptance probability makes the first-order condition precisely the reduced-load equations above. Conversely every fixed point is such a minimum. Thus fixed routing has exactly one Erlang fixed point, including zero blocking at unused resources. The proof does not rely on a potentially nonconvergent simultaneous substitution algorithm.

Take a finite set of source-sink pairs $S$, with fixed demands $b_s\ge 0$, and finitely many permitted routes $R_s$ for each pair. Every positive demand must have at least one available route. Let ${\nu }_r$ be route flow and ${\rho }_j$ link throughput. The source-sink route incidence matrix and link-route incidence matrix arefor simple routes; if a route traverses a link repeatedly, its entry counts those traversals. Thus $H\nu =b$ conserves each demand and $A\nu =\rho$ aggregates link traffic.

At a Wardrop equilibrium, every route carrying positive flow has minimum delay among routes for the same source-sink pair. Ifthere are numbers ${\tau }_s$ such that $c_r(\rho )\ge {\tau }_s$ for $r\in R_s$, with equality whenever ${\nu }_r>0$. This is a nonatomic routing game: one infinitesimal user's route change does not alter the link costs it sees.

For continuous nondecreasing link delays, define the Beckmann potentialEach term is convex because its derivative is nondecreasing. The feasible route-flow set $\mathcal{F}=\{\nu \ge 0:H\nu =b\}$ is nonempty and compact: $0\le {\nu }_r\le b_s$ on each pair's routes. Hence the continuous function $V(A\nu )$ attains a minimum. Its gradient with respect to route flow isIf $r$ carries positive flow and $r^{\prime }$ serves the same pair, moving a small amount of flow from $r$ to $r^{\prime }$ is feasible. At a minimizer its directional derivative $c_{r^{\prime }}-c_r$ is nonnegative. Thus every used route minimizes delay, proving existence of a Wardrop equilibrium.

Conversely, at such an equilibrium, any feasible competitor $\tilde{\nu }$ satisfiesIndeed, the competitor's cost sum for pair $s$ is at least ${\tau }_s{b}_s$, while the current flow's cost sum equals ${\tau }_s{b}_s$. This first-order variational inequality is sufficient for optimality of a differentiable convex function. ThereforeThe argument proves both directions of the claimed optimization equivalence, including unused routes with costs tied at the minimum.

If each $D_j$ is strictly increasing, each integral is strictly convex, hence so is $V$ as a function of link throughputs. The feasible link-throughput set $A\mathcal{F}$ is convex. Two distinct minimizing vectors would have a feasible midpoint with strictly smaller potential, so the equilibrium link-throughput vector is unique. Strictly convex function in $\rho$ need not give strictly convex function in $\nu$, because $A$ can have a nontrivial kernel.

For an explicit route-flow ambiguity, put two parallel links in each of two consecutive stages, with one unit of demand traversing both stages. Order the four routes by the first and second link choices: $(1,1),(1,2),(2,1),(2,2)$. Set all four link delays to $D_j(z)=z$. Theninduces throughput $1/2$ on each link, and every route has delay one. Every member of this family is an equilibrium, though its route-flow vector changes. This is route-flow nonuniqueness at a Wardrop equilibrium with strictly increasing delays.

For Braess paradox, use vertices $s,u,v,t$ and one unit of $s$-to-$t$ demand. The delays are $D_{su}(z)=z$, $D_{ut}(z)=1$, $D_{sv}(z)=1$, and $D_{vt}(z)=z$. Initially the only routes are $sut$ and $svt$. If the upper route carries $p$, their costs are $1+p$ and $2-p$, so equilibrium has $p=1/2$ and common delay $3/2$.

Add the directed link $u\to v$ with delay zero. Let the upper, lower and new middle routes carry $p,q,c$, where $p+q+c=1$. Their costs becomeIf $p>0$, the middle route is strictly cheaper than the upper route, contradicting equilibrium; likewise $q>0$ is impossible. Thus all flow uses the middle route, with $p=q=0,c=1$, and all three route costs are two. Therefore

The potential for the expanded network isThe old flow remains feasible and has potential $5/4$; the new equilibrium has potential $1$. The enlarged feasible set has indeed lowered the minimized potential. But total travel delay is ${\sum }_j{\rho }_j{D}_j({\rho }_j)$, not this integral potential: it rises from $3/2$ to $2$. This explains the paradox within the optimization framework. The optimal total-delay value cannot worsen when the old flow is still allowed; selfish equilibrium minimizes a different function.

The phenomenon also survives strictly increasing delays on every link. Replace each constant-one delay by $1+\epsilon z$ and the new-link delay by $\epsilon z$. Symmetry gives upper and lower flows $p=\epsilon /(1+3\epsilon )$ after expansion, with middle flow $(1+\epsilon )/(1+3\epsilon )$. The common new delay is $2-\epsilon (1-\epsilon )/(1+3\epsilon )$, while the old delay was $3/2+\epsilon /2$. For $\epsilon =1/10$, these are $251/130>31/20$. Thus the example does not depend on interpreting increasing delays as merely nondecreasing.

For active-route per-flow rates $x_r>0$, feasibility means ${\sum }_{r:j\in r}{n}_r{x}_r\le C_j$ for every resource. The defining condition for a proportionally fair allocation is that every feasible alternative $\tilde{x}$ satisfiesThis is equivalent to maximizing weighted logarithmic utility ${\sum }_{r:n_r>0}{n}_r\log x_r$ over the feasible set. Its gradient is $(n_r/x_r)$, so the displayed inequality is precisely the first-order condition for this concave function. Strict concavity gives unique active-flow rates. Routes with $n_r=0$ have no utility term or actual per-flow rate to determine; set their aggregate service to zero. This is the inactive routes in proportional fairness convention.

Abbreviate the route populations as $n_1,n_2,n_3,n_4$ in cyclic order and let $z_r=n_r{x}_r$ denote aggregate route service. The resource constraints areEquivalently, $\max (z_1,z_3)+\max (z_2,z_4)\le 1$. Fix these two maxima as $u,v$. Because utility is increasing in each active allocation, every active odd route attains $u$ and every active even route attains $v$. Dropping terms constant in the populations, the objective becomesIf both groups are nonempty, the maximum has $u+v=1$ and derivative $P/u-Q/(1-u)=0$, giving $u=P/M,v=Q/M$. If one group is empty, every route in the other group gets aggregate service one. For $M>0$, all cases are summarized, on active routes, byAt $M=0$ there are no flows and all aggregate services are zero. In particular $n_1{x}_1=(n_1+n_3)/M$ whenever $n_1>0$, as printed in the PDF. The per-flow rate is not generally $1/M$: opposite routes can use the same resource capacities in parallel.

An exponential document size with parameter ${\mu }_r$ has completion hazard ${\mu }_r{x}_r$ when a flow is served at rate $x_r$. Summing over its $n_r$ flows gives the transition ratesThus an active odd route departs at rate ${\mu }_{r}P/M$, and an active even route at rate ${\mu }_{r}Q/M$. An empty route has departure rate zero even if its opposite route is populated.

Put ${\alpha }_r={\nu }_r/{\mu }_r$ andFor an active odd route and an active even route respectively,Hence aggregate service is the neighboring-state ratio of this balance function of a flow-level network. Consider weights $w(n)=\Phi (n){\prod }_r{\alpha }_r^{n_r}$. For every birth edge,These are detailed balance equations. The total departure rate is bounded by ${\sum }_r{\mu }_r$ and total arrival rate by ${\sum }_r{\nu }_r$, so the process is nonexplosive. Provided the normalizing sum is finite, its stationary distribution is thereforeThe large parentheses in the original PDF represent a binomial coefficient, not a ratio. The verified adjacent-state ratios explain why that distinction matters.

The required stability condition can be established directly from the normalizing sum. Set $a={\alpha }_1,c={\alpha }_3,b={\alpha }_2,d={\alpha }_4$, and $h_p(a,c)={\sum }_{i=0}^p{a}^i{c}^{p-i}$. Grouping opposite-route counts givesIf $A_{\ast }=\max (a,c)$ and $B_{\ast }=\max (b,d)$, then $A_{\ast }^p\le h_p\le (p+1)A_{\ast }^p$, with the same bounds for the other group. Restriction to the routes attaining the maxima bounds $B$ below by ${\sum }_m(A_{\ast }+B_{\ast }{)}^m$. If $A_{\ast }+B_{\ast }<1$, the upper bound is at most ${\sum }_m(m+1{)}^2(A_{\ast }+B_{\ast }{)}^m$, which converges. ThusEquivalently every resource's offered load is strictly below its unit capacity. With positive arrival rates the irreducible chain then has the unique stationary probability distribution above. Without this condition the weights remain an invariant measure but are not normalizable; the question's probability-distribution claim implicitly needs this hypothesis.

For completeness, the normalizing constant has the four-cycle partition function expressionTo derive it when $a\ne c$ and $b\ne d$, substitute $h_p=(a^{p+1}-c^{p+1})/(a-c)$ into the double sum and use ${\sum }_{p,q}\left(\genfrac{}{}{0ex}{}{p+q}{p}\right)u^p{v}^q=(1-u-v{)}^{-1}$. The resulting expression iswhich simplifies to the displayed formula. Continuity within the stability region covers equal opposite loads and zero loads.

The third planform is fixed by a diagonal reflection and changes sign under a quarter-turn. Extend the preceding group representation byThe other diagonal reflection changes only the sign of $B$. Consequently the first two components of an equivariant dynamical system have factors $A$ and $B$, whereas its third component is even in each of $A,B$. The quarter-turn exchanges $A^2$ and $B^2$ while negating $C$. Its admissible quadratic coupling is therefore $A^2-B^2$ in the third equation and opposite $AC,BC$ couplings in the first two. At cubic order the only new terms are $A{C}^2,B{C}^2,C^3,C(A^2+B^2)$. The linear action on the two-dimensional irreducible mode has one coefficient ${\mu }_1$, while the distinct scalar representation has its own coefficient ${\mu }_2$. A small splitting of the two marginal linear modes supplies these two unfolding parameters. With real coefficients the resulting amplitude equations areThese equations follow by listing all equivariant monomials through degree three; the displayed minus signs are coefficient conventions, not consequences of symmetry.

In the specified reduced system put $A=B=s$, $C=0$. A nonzero branch exists for ${\mu }_1>0$, with $q=s^2={\mu }_1/(1+\lambda )$. Write perturbations as $u=(\delta A+\delta B)/2$, $v=(\delta A-\delta B)/2$, and $c=\delta C$. Linearization givesFor the square-pattern interaction with a sign-changing scalar mode, the remaining eigenvalues have sum $\tau =p+{\mu }_2$ and product $D=p{\mu }_2+4{\alpha }_1{\gamma }_{1}q=p({\mu }_2+K)$, where $K=2{\alpha }_1{\gamma }_1/(\lambda -1)>0$. A real two-dimensional stability matrix has both eigenvalues in the left half-plane exactly when its trace is negative and its determinant is positive. ThereforeIf ${\mu }_2<-K$, the block has negative determinant and one growing direction, explaining the instability at sufficiently small ${\mu }_2$. At the lower endpoint ${\mu }_2=-K$, one eigenvalue is zero and the other is $p-K<0$. The symmetry $(A,B,C)\mapsto (B,A,-C)$ fixes the diagonal state and negates its critical $(v,c)$ direction. Thus a generic nonlinear unfolding has a pitchfork bifurcation to unequal-amplitude states with nonzero $C$.

At the upper endpoint ${\mu }_2=-p$, the trace vanishes andThe symmetric eigenvalue remains negative. The critical conjugate pair crosses the imaginary axis with derivative $d\mathrm{Re}\sigma /d{\mu }_2=1/2$. If its cubic Hopf normal form coefficient is nonzero, this is a Hopf bifurcation and nearby periodic solutions exist on one side. Their stability and the side of existence depend on that coefficient; the linear calculation does not determine them. It does establish the generic possibility of oscillating square patterns. The equality $p=K$ is a double-zero degeneracy, outside this nondegenerate Hopf bifurcation argument.

Write $s=R^2$. A steady complex amplitude obeys $(\mu -s+i\Lambda )A=-1$, soThus every positive intensity root gives exactly one steady phase, and $s=0$ is impossible. The cubic intensity polynomial is $F(s,\mu )=s^3-2\mu s^2+({\mu }^2+{\Lambda }^2)s-1$. Its positive steady branch can be parameterized completely asThe two formulae join smoothly as a single curve at the maximum radius $R={\Lambda }^{-1}$; this joining point is not a saddle-node bifurcation. A turning point with respect to $\mu$ satisfies $F_s=0$. Subtracting the equilibrium and derivative conditions givesThese points lie on ${\mu }_{+}$. Since $f^{\prime }(s)=(1-s^3)/s^5$, its maximum is $f(1)=3/4$. Therefore there is exactly one distinct steady state for every $\mu$ when $\Lambda \ge \sqrt{3}/2$. One can also see uniqueness directly: ${\mu }_{-}$ is increasing and ${\mu }_{+}$ is decreasing when no turning points exist, and their respective ranges meet only at their common endpoint. At the equality there is a cusp bifurcation with a triple intensity root at $s=1,\mu =3/2$.

When $0<\Lambda <\sqrt{3}/2$, the two turning intensities satisfy $s_{-}<1<s_{+}$. Define ${\mu }_L=s_{-}+1/(2s_{-}^2)$ and ${\mu }_U=s_{+}+1/(2s_{+}^2)$; the curve has a local minimum at the former and a local maximum at the latter, with ${\mu }_L<{\mu }_U$. There are three distinct steady intensities for ${\mu }_L<\mu <{\mu }_U$, two distinct ones at an endpoint, and one otherwise. The bifurcation diagrams below show the folded small-$\Lambda$ curves and the unfurled large-$\Lambda$ curve, together with the linear stability calculated next. Both ends of the full branch have $R\to 0$ as $|\mu |\to \infty$, while its middle reaches the finite maximum ${\Lambda }^{-1}$.

Let the uniform nonzero amplitude be $R$, so $\mu =R^2-\alpha R$ and $B=0$. Write $r=R(2R-\alpha )$. A perturbation with Fourier wavenumber $k=n\pi /L$ has stability matrixThe constant $B$ mode is disallowed because the conserved mean is prescribed. At $k=0$ the only allowed mode has eigenvalue $-r$, so homogeneous stability requires $r>0$. Then every nonzero-mode trace is negative, and its determinant is smallest in sign at the smallest allowed $k^2$. The exact periodic stability criterion for a conserved-field uniform amplitude isA zero in either inequality is a marginal case requiring nonlinear analysis. The zero-amplitude branch, separately, has amplitude growth rates $\mu -k^2$ and nonzero conserved-field growth rates $-\sigma k^2$, and is strictly stable when $\mu <0$.