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Assume global four-dimensional N=1 supersymmetry with canonical kinetic terms and no explicit gaugino mass. Write $W_i={\partial }_{i}W$ and $W_{ij}={\partial }_i{\partial }_{j}W$. To match the displayed stationarity equation, use $F^i=\overline{W_i}$; this differs by a minus sign from the eliminated auxiliary field in the component convention of Solution 2. The real D-terms are $D^a=-g^a({\bar{\varphi }}_i(T^a{)}^i{}_j{\varphi }^j+{\xi }^a)$, with Hermitian gauge generators. A constant Fayet–Iliopoulos term is permitted only for an appropriate Abelian factor. The scalar potential isDifferentiate with respect to ${\varphi }^i$, holding $\bar{\varphi }$ independent:Thus stationary point conditions gives the first requested identity. Gauge invariance of the superpotential gives $W_i(T^a{)}^i{}_j{\varphi }^j=0$. Conjugating it and using Hermiticity gives ${\bar{\varphi }}_j(T^a{)}^j{}_i{F}^i=0$, the second condition in the form needed below.

At the vacuum, define $H_{ij}=⟨W_{ij}⟩$ and $G_{ia}=g^a⟨{\bar{\varphi }}_j⟩(T^a{)}^j{}_i$. The two conditions combine asHere the upper equation is stationarity and the lower equation is conjugated gauge invariance. The chiral-superfield fermion mass matrix contributes $-\frac{1}{2}{H}_{ij}{\psi }^i{\psi }^j$, and the gauge Yukawa interaction contributes the mixed chiral-fermion/gaugino bilinear. With canonical Weyl fermion and gaugino kinetic terms, and choosing the gaugino phase to put the mixed bilinear in real-sign convention, these terms areThe factor $\sqrt{2}$ is important: the matrix directly acting on the printed $(F,D)$ is the same mass bilinear with rescaled gauginos, rather than literally the canonical matrix. In detail, for $S=\mathrm{diag}(I,\sqrt{2}I)$,Since supersymmetry breaking makes this vector nonzero, there is at least one zero mass eigenvalue. A normalized massless fermion isWith the consistent gaugino phase, the inhomogeneous vacuum supersymmetry transformation of $(\psi ,\lambda )$ is proportional to the same vector $(F,D/\sqrt{2})$. Therefore $\eta$ shifts by a nonzero constant times the transformation parameter: it is the Goldstino, the Goldstone fermion of broken global supersymmetry. The argument guarantees at least one Goldstino zero mode, not the absence of other massless fermions.

For the second calculation, first take only canonical chiral superfields and F-term breaking. At a stationary vacuum $v$, let $h=\varphi -v$, $m_{ij}=W_{ij}(v)$ andThese follow by differentiating $V={\sum }_k|W_k{|}^2$ twice. The quadratic scalar potential isThe doubled scalar mass matrix has blocksTransforming to the $2N$ canonically normalized real components of $h$ preserves its trace, so the sum of real scalar mass squares is $\mathrm{tr}{M}_s^2=2\mathrm{tr}A$. The Weyl fermion mass squares are the eigenvalues of $m^{†}m=A$ and their sum is $\mathrm{tr}A$. Each Weyl mass eigenstate has two physical spin states. The printed supertrace has scalar sign $-1$ and fermion weight $+2$, henceThe frequently used boson-minus-fermion definition has the opposite overall sign and gives the same zero. The mixing block $B$ can split scalar masses but cannot change this trace. This proves the tree-level supertrace mass sum rule; it uses canonical kinetic terms and a renormalizable global theory, rather than a general supergravity or noncanonical effective theory.

In the MSSM, trying to make all squarks and sleptons heavy through direct canonical tree-level F-term breaking encounters this constraint. In an unbroken electric-charge and color sector, the scalar mass-squared trace is twice the trace for its ordinary fermionic partners. More explicitly, take a unit vector $u$ minimizing $u^{†}Au$. For $h=e^{i\alpha }u$, choose $\alpha$ so that $\mathrm{Re}(e^{2i\alpha }{u}^{T}Bu)\le 0$. The curvature along this real scalar direction is then at most $u^{†}Au$, so at least one scalar mass square is no larger than the smallest fermion mass square in that sector. For one isolated pair the scalar masses are $m_f^2\pm |B|$. Thus light quarks can force an unacceptably light squark, or a tachyon, under these assumptions. A hidden supersymmetry-breaking sector with mediation produces effective soft supersymmetry breaking; radiative effects, noncanonical kinetic terms, or supergravity can change the assumptions and allow a viable spectrum. The trace over an arbitrary enlarged theory alone would not locate the light state in the visible sector; the MSSM conclusion uses the conserved-charge block structure.

The first alternative in the supplied analytic-group theorem would give a faithful representation of $\mathcal{N}$ in some ${\mathrm{GL}}_d({\mathbb{Z}}_p)$ or ${\mathrm{GL}}_d({\mathbb{F}}_p[[t]])$. The embedding proved above supplies cyclic subgroups of order $p^a$ for arbitrarily large $a$. Such orders are impossible in either fixed matrix dimension.

In characteristic zero, a matrix $A$ of order $p^a$ is diagonalizable over an algebraic closure, because its minimal polynomial divides the separable polynomial $X^{p^a}-1$. At least one eigenvalue must be a primitive $p^a$th root of unity; otherwise $A^{p^{a-1}}=I$. The cyclotomic polynomial ${\Phi }_{p^a}$ is irreducible over ${\mathbb{Q}}_p$, since ${\Phi }_{p^a}(1+X)$ satisfies the Eisenstein criterion at $p$. Consequently that eigenvalue has degree $\phi (p^a)=p^{a-1}(p-1)$ over ${\mathbb{Q}}_p$, which must be at most the matrix dimension $d$. This bounds $a$.

In characteristic $p$, pass from the power-series ring to its fraction field. A matrix of $p$-power order satisfies $X^{p^a}-1=(X-1{)}^{p^a}$, so it is unipotent. Writing $A=I+B$, nilpotence gives $B^d=0$. For $p^e\ge d$,Thus its order is at most $p^{\lceil {\log }_{p}d\rceil }$, again a fixed bound. Therefore the linear alternative is impossible in both coefficient rings.

Write $S={\Phi }_0$, $u={\Phi }_{+}$, $v={\Phi }_{-}$, $M=M_p$, $P=uv-\zeta$, $x=|S{|}^2/M^2$ and $q=|u{|}^2+|v{|}^2$. Assume $g\ne 0$ and, initially, $\zeta \ne 0$. The supergravity F-term potential isThe canonical Kähler potential gives the identity Kähler metric. The three Kähler covariant derivatives of a superpotential areSubstituting, including the negative term rather than treating the potential as just a sum of squares, gives the exact answerHere $(1+x{)}^2-3x=1-x+x^2$ explains the first coefficient. It is strictly positive for real $x$, since $1-x+x^2=(x-1/2{)}^2+3/4$. Thus this model's scalar potential is nonnegative everywhere.

In the stated limit $q/M^2\ll 1$, the exponential can be replaced by $e^x[1+O(q/M^2)]$. The other displayed terms retain the effects of the Kähler covariant derivatives of a superpotential. Small waterfall fields alone do not justify discarding $\zeta /M^2$ in those derivatives, or expanding in $x$. If also $|\zeta |/M^2\ll 1$ and $x\ll 1$, the leading result reduces to the global F-term scalar potentialNo gauge sector or gauge coupling is supplied, so there is no specified D-term contribution to add.

The supersymmetric vacuum manifold of a bilinear hybrid superpotential isIndeed $W=0$ and all three Kähler covariant derivatives of a superpotential vanish there, so all supergravity auxiliary fields vanish and supersymmetry is unbroken. The nonnegative scalar potential makes this a global minimum. Conversely zero potential requires $uv=\zeta$, and then $|S{|}^{2}q=0$; for $\zeta \ne 0$ this forces $S=0$. With just the supplied fields this is a complex vacuum manifold before imposing any gauge quotient: the product is fixed but the ratio of $u$ and $v$ is not. More precisely, it is an F-flat vacuum manifold whether or not a gauge symmetry is specified. If one additionally gauges opposite charges, D-flatness typically imposes $|u|=|v|$, giving $|u|=|v|=\sqrt{|\zeta |}$ up to phases and gauge equivalence; that is an extra assumption, not a consequence of $K$ and $W$ alone.

The other familiar branch of F-term hybrid inflation has $u=v=0$. Its nonzero derivative $D_{S}W=-g\zeta (1+x)$ establishes supersymmetry breaking, and its exact scalar potential isThe canonical-Kähler correction to an F-term hybrid valley begins asIn particular the inflaton quadratic mass cancels at this order; the first lift of the global flat valley is quartic in $|S|$.

To identify when this branch is a transverse minimum, rephase the waterfall fields so that $\zeta$ is real positive. To quadratic order in $u,v$, the global scalar potential isThe normalized combinations $(u\pm \bar{v})/\sqrt{2}$ diagonalize this quadratic form, withThus $|S{|}^2>|\zeta |$ gives a valley of transverse minima, while below the threshold the negative eigenvalue produces the waterfall instability in F-term hybrid inflation.

The exact supergravity correction to this stability test is also available without taking $x$ small. Put $b=|\zeta |/M^2$. Expansion of the exact supergravity F-term potential about $u=v=0$ gives equal diagonal coefficients $|g{|}^2{M}^2{e}^x[x+b^2(1+x^2)]$ and off-diagonal magnitude $|g{|}^2{M}^2{e}^{x}b(1+x+x^2)$. HenceFor the usual sub-Planckian symmetry-breaking scale $0<b\ll 1$, the lower eigenvalue is $|g{|}^2{M}^2{e}^x(1-b)[x-b(1+x^2)]$. It is positive for the familiar sub-Planckian valley $b\ll x\ll 1$, and changes sign near $x=b$.

The printed assertion of two full minima requires qualification. The broken branch is a transverse inflationary valley, not generally a second stationary minimum of the complete supergravity potential. In factIts only stationary point in $S$ is the origin. For $0<|\zeta |<M^2$ that point has a negative waterfall eigenvalue $|g{|}^2(|\zeta {|}^2/M^2-|\zeta |)$. A direct counterexample is the path $S=0$, $u=v=a$ with real small $a$, after rephasing $\zeta >0$:The negative quadratic coefficient proves it is not a minimum. No loop corrections or additional stabilizing interactions were specified that could create the requested second stationary vacuum. The mathematically supported interpretation is a SUSY-breaking transverse valley and a zero-energy supersymmetric vacuum minimum manifold. The limit on $u,v$ alone does not fix the parameter range $|\zeta |/M^2$; the familiar waterfall conclusion uses the sub-Planckian range just stated.

In the corresponding global theory the exponential prefactor, Kähler corrections to $D_{i}W$ and universal $-3|W{|}^2/M^2$ term are absent. Its $u=v=0$ valley has exactly constant tree-level energy $|g\zeta {|}^2$ for all $S$, whereas canonical supergravity lifts it by the quartic correction above. The supersymmetric vacuum $S=0$, $uv=\zeta$ and its zero energy persist in both descriptions. If $\zeta =0$, the origin and the $S$ axis instead have vanishing F-terms; the broken-valley interpretation does not apply.

Write $a=⟨\psi |w⟩=2^{-N/2}$, which is real and satisfies $0<a<1$ for $N\ge 1$. The invariant two-dimensional span has orthonormal eigenvectorswith eigenvalues $\Delta (1+a)$ and $\Delta (1-a)$. In this basis,Transfer therefore requires the two evolved eigenphases to differ by an odd multiple of $\pi$:The first positive time isEquivalently, exponentiating on the invariant span gives the full evolutionAt $t_0$ this is $-i{e}^{-i\Delta t_0}|w⟩$, so the requested global phase can be chosen as $\alpha =-\Delta t_0-\pi /2$ modulo $2\pi$. Since the two vectors are linearly independent, their relative coefficient cannot produce exact transfer at an earlier positive time. This is continuous-time quantum search, with time proportional to the square root of the search-space dimension for fixed $\Delta$. Restoring units multiplies the stated time by $\hslash$. The degenerate zero-qubit case would already have $\psi =w$, so there would be no smallest positive transfer time; the formula concerns the nontrivial $N\ge 1$ search.

Write $|0⟩=|\uparrow ⟩$, $|1⟩=|\downarrow ⟩$, and use $X_j,Y_j$ for the Pauli matrices on particle $j$. Their actions on the computational basis areSet $P_1=X_1{Y}_2{Y}_3$, $P_2=Y_1{X}_2{Y}_3$, $P_3=Y_1{Y}_2{X}_3$, and $P_4=X_1{X}_2{X}_3$. Any two distinct operators in this list differ by an $X$ versus $Y$ at exactly two sites. At each such site $XY=-YX$, while factors on different particles commute. Swapping the two full products therefore introduces two minus signs, givingThis is the Pauli-string commutation parity rule. Each product is Hermitian and squares to the identity, so its only eigenvalues are $\pm 1$.

For the negative-phase Greenberger–Horne–Zeilinger state, put $|G_{-}⟩=(|000⟩-|111⟩)/\sqrt{2}$. The two $Y$ factors in each of $P_1,P_2,P_3$ contribute $i^2=-1$ on $|000⟩$ and $(-i{)}^2=-1$ on $|111⟩$. Thus these operators take $|000⟩$ to $-|111⟩$ and $|111⟩$ to $-|000⟩$. By contrast, $P_4$ flips these two basis states without a phase. ConsequentlyThe GHZ state is therefore a simultaneous eigenstate of the four commuting products. As a sign check, direct multiplication givesThis is compatible with the displayed quantum eigenvalues.

Now apply the EPR criterion of reality together with locality for the spacelike separated particles. In the context with local settings $X_1,Y_2,Y_3$, the product of the three outcomes is certainly $+1$. Therefore the two remote $Y$ outcomes predict the first particle's $X$ outcome with certainty, without disturbing it under the locality assumption. The other perfect-correlation contexts similarly predict each particle's $X$ or $Y$ outcome from measurements performed only on the other particles. The EPR criterion of reality then assigns predetermined local values $x_j,y_j\in \{+1,-1\}$ to these six observables. Locality makes the value assigned at one site independent of which measurements were chosen at the other sites.

To reproduce the three certain mixed-setting quantum predictions, every such assignment must satisfyMultiplying these scalar equations cancels each squared $y_j$ and gives $x_1{x}_2{x}_3=+1$. But the all-$X$ quantum prediction, from the $P_4$ eigenvalue, is $x_1{x}_2{x}_3=-1$ with certainty. HenceThe perfect GHZ correlations contradict the simultaneous local elements of reality inferred by the EPR argument. No statistical Bell inequality is needed. The locality assumption is essential to that inference; the certainty criterion without the premise that distant choices cannot disturb the local element does not by itself imply a local hidden-variable model. Nor does the argument require simultaneous measurement of $X$ and $Y$ on a single particle: different experimental contexts establish the perfect predictions, and it is the EPR/locality interpretation that attempts to assign all their values at once.

Take any rank-one orthogonal projection $P$ and its generalized Bloch vector $s$. The antipodal vector $-s$ is on the same sphere, but its corresponding trace-one Hermitian matrix isOn the range of $P$, this matrix has eigenvalue $2/N-1$, which is negative for $N>2$. It is therefore not a positive semidefinite matrix and cannot be a density operator. Hence the pure states do not cover the sphere when $N>2$. This also shows directly why the purity bound for generalized Bloch vectors does not by itself guarantee a physical quantum state.

The eigenvalues $p_j$ of a density operator are nonnegative and sum to one. ThereforeFor a non-pure state at least two eigenvalues are positive, so the last expression is strictly positive. Combining this with the Generalized Bloch representation givesThus every non-pure quantum state lies strictly inside the radius sphere for $N\ge 2$. It need not be an interior point of the entire physical state body: for $N>2$, a rank-deficient mixed density operator can lie on that body's boundary while still lying inside the Euclidean sphere.

In units $\hslash =1$, a bilinear Hamiltonian control system haswhere the Hermitian matrices $H_0,H_k$ are fixed, and the real functions $u_k$ are admissible control inputs. The corresponding vector or density operator dynamics are $\dot{\psi }=-iH(u)\psi$ or $\dot{\rho }=-i[H(u),\rho ]$. The dynamics are linear in the state at fixed input and affine in the inputs at fixed state; this is the bilinear terminology.

The reachable set from an initial state $x_0$ at time $T$ is ${\mathcal{R}}_T(x_0)=\{x(T;x_0,u):u\text{ admissible}\}$. With freely chosen duration, take $\mathcal{R}(x_0)={\bigcup }_{T\ge 0}{\mathcal{R}}_T(x_0)$. Controllability means every allowed target is reachable from every allowed initial state. The allowed space and phase convention must be specified. For closed quantum control, density operators remain on their initial unitary orbit of a density operator, rather than becoming arbitrary density operators with different eigenvalues.

The two density operators have the same eigenvalues $a,a,b,b$, so a permutation of basis vectors makes them unitarily equivalent. A dynamical equivalence must additionally respect the symplectic dynamical symmetry in quantum control.

Define the symplectic dual $\tilde{\rho }=J{\rho }^T{J}^{†}$. For a unitary $U$ preserving $J$, both $U^{T}JU=J$ and $UJ{U}^T=J$ hold. In particular $J{U}^{\ast }=UJ$, and thereforeIt follows by cyclicity of the trace that the symplectic invariant of a density operatoris unchanged by every reachable unitary conjugation. The given $J$ reverses the diagonal order, soConsequentlyHence the states are not dynamically equivalent, despite their identical ordinary spectra. This works for all the allowed values, including $a=0$, and for any subgroup satisfying the symmetry in part (c).

For an alpha-quenching rotating wave, write the four complex fields as constant amplitudes times $e^{i\omega t}$, with $\omega$ real. The squared toroidal amplitudes are then constant, so the algebraic alpha quenching coefficients are constant on the orbit. Assume $\Omega \ne 0$. With ${ϵ}_2=0$, the toroidal equations determineSubstitution in the potential equations gives the amplitude eigenvalue problemThe matrix is real symmetric, hence Hermitian, and every one of its eigenvalues is real. For a nonzero amplitude vector this forces $1-{\omega }^2=0$. Thus every nonzero rotating wave of this form hasSet $a={\alpha }_0-2\omega /\Omega$. In the dipole subspace $B_2=-B_1$ and $A_2=A_1$; in the quadrupole subspace $B_2=B_1$ and $A_2=-A_1$. The amplitude equations giveThese are nonzero waves exactly when the respective squared amplitude is positive. A common constant phase of $B_1,B_2$ is arbitrary, and $|A_j{|}^2=2|B_j{|}^2/{\Omega }^2$. In particular, choosing $\omega =\mathrm{sgn}\Omega$, both parities exist once ${\alpha }_0>2/|\Omega |+|{ϵ}_1|$. The other frequency sign is also algebraically possible whenever its positivity conditions hold; existence alone says nothing about stability of that branch.

There is a genuine sign qualification in the polynomial alpha quenching law. It subtracts $p|B{|}^2$ independently of the sign of ${\alpha }_0$. Thus a sufficiently large positive ${\alpha }_0$ gives the intended waves, but a sufficiently large negative ${\alpha }_0$ makes every displayed squared amplitude negative for both frequency signs. For example, with $\Omega =1$, ${ϵ}_1=0$, ${\alpha }_0=-M$ and $M>2$, neither $(-M-2)/p$ nor $(-M+2)/p$ is nonnegative. Large $|{\alpha }_0\Omega |$ alone is insufficient as printed; positive alpha, or the exact amplitude-positivity conditions above, is required. A sign-preserving quenching prescription would be a different model.

For the unequal-hemisphere dynamo rotating wave, first take ${ϵ}_1\ne 0$. The real symmetric matrix above has distinct eigenvalues because its off-diagonal entry is nonzero. Its eigenvector is real up to a common complex phase, so choose $B_1=b_1$, $B_2=b_2$ real. The amplitude equations reduce toMultiplying the first equation by $b_2$, the second by $b_1$, and subtracting givesFor unequal magnitudes, therefore, $b_1{b}_2=-{ϵ}_1/p$. Neither amplitude can vanish. Multiplying the first amplitude equation by $b_1$ then shows $ap{b}_1^2-p^2{b}_1^4-{ϵ}_1^2=0$. Equivalently,The two distinct positive roots exist precisely when $a>2|{ϵ}_1|$. This constructs the desired solutions explicitly:Choose $b_1$ to be either square root of the first value, and set $b_2=-{ϵ}_1/(p{b}_1)$; substituting verifies both amplitude equations. Interchanging the two roots gives the hemisphere-exchanged companion. The hinted choice $\omega =1$ therefore requires ${\alpha }_0-2/\Omega >2|{ϵ}_1|$. At equality the magnitudes coincide, so it is not an asymmetric solution. For ${ϵ}_1>0$ the amplitudes have opposite signs and meet the dipole branch at equality; for ${ϵ}_1<0$ they have the same sign and meet the quadrupole branch.

If ${ϵ}_1=0$, the hemispheres decouple. An asymmetric rotating wave instead has one hemisphere inactive and the other with $|B{|}^2=a/p>0$. If both hemispheres are active their magnitudes must be equal, although their relative phase is arbitrary. If $\Omega =0$, the toroidal equations simply give ${\dot{B}}_j=-B_j$, so no nonzero common-frequency rotating wave exists.

The unequal-amplitude solutions describe a periodically reversing dynamo concentrated more strongly in one hemisphere, with $|A_1|\ne |A_2|$ following from $|A_j|=\sqrt{2}|B_j|/|\Omega |$. Both $B_d=(B_1-B_2)/2$ and $B_q=(B_1+B_2)/2$ are nonzero, so the field combines the two dipole and quadrupole parity in a mean-field dynamo classes rather than belonging to either one. Equatorial reflection exchanges the two companion solutions. The model can therefore break hemispheric symmetry without asymmetric input parameters. This construction proves existence; nonlinear stability would require an additional perturbation calculation.

Take the coupling parameters to be real. The dipole and quadrupole parity in a mean-field dynamo subspaces are invariant: substitution of $A_2=A_1$, $B_2=-B_1$ makes the second-hemisphere equations equal to the first-hemisphere equations with these same signs. Substitution of $A_2=-A_1$, $B_2=B_1$ likewise preserves the other parity. In either subspace put $A=A_1$, $B=B_1$. The reduced equations and characteristic equations areA marginal normal mode has $\lambda =i\omega$. Equating real and imaginary parts givesConsequently the marginal surfaces areTo identify which side actually grows, write $z=ab$ in $(\lambda +1{)}^2=z$. The largest real part of the two eigenvalues is $-1+\sqrt{(|z|+\mathrm{Re}z)/2}$. It is negative exactly when $\mathrm{Re}z+[\mathrm{Im}z{]}^2/4<1$: squaring $|z|<2-\mathrm{Re}z$ proves this, and the latter condition ensures the right-hand side is positive. Thus dipole modes decay for $F_d<1$ and quadrupole modes decay for $F_q<1$, with growth for the corresponding reversed inequality.

For explicit coupled-hemisphere dynamo parity thresholds, suppose $\Omega \ne 0$ and defineThe dipole stability interval is $\delta -L<{\alpha }_0<\delta +L$, and the quadrupole stability interval is $-\delta -L<{\alpha }_0<-\delta +L$. A comparison of first onset from ${\alpha }_0=0$ assumes both are initially stable, namelyIncrease $|{\alpha }_0|$ along a specified sign $s=\mathrm{sgn}{\alpha }_0$. The first thresholds for the two parities areTherefore the quadrupole reaches onset first if $s({\Omega }^2{ϵ}_1+2{ϵ}_2)>0$; the dipole reaches onset first if it is negative; the thresholds coincide if it vanishes. For positive ${\alpha }_0$ and positive coupling parameters this selects the quadrupole, provided the initial stability condition holds. Reversing the sign of ${\alpha }_0$ exchanges the preference. If both signs are freely allowed, each parity attains the same minimum threshold $|\Omega |(L-|\delta |)$ on opposite sign rays; a unique preference cannot be inferred from the absolute product alone. If the initial stability condition fails, the zero-alpha state is already unstable or marginal and there is no first onset from a stable origin. When $\Omega =0$, the formulas for $F_d,F_q$ still apply, but $|{\alpha }_0\Omega |$ is identically zero and is not a useful onset parameter.

Fourier stability analysis exploits translation invariance of a linear constant-coefficient finite difference method. On an infinite uniform grid of spacing $h$, a spatial shift acts on $e^{ij\theta }$ by multiplication by $e^{i\theta }$. Hence a scalar two-level scheme with new- and old-time symbols $L(\theta ),N(\theta )$ propagates each Fourier mode by $G(\theta )=N(\theta )/L(\theta )$, provided the new-time operator is invertible. An explicit stencil $U_j^{n+1}={\sum }_{\ell }{a}_{\ell }{U}_{j+\ell }^n$ simply has $G(\theta )={\sum }_{\ell }{a}_{\ell }{e}^{i\ell \theta }$. This converts a spatial evolution into scalar algebra at each frequency.

The Plancherel theorem gives the full norm estimate, rather than just a heuristic about waves. For the discrete Fourier transform $\hat{U}(\theta )={\sum }_j{U}_j{e}^{-ij\theta }$, interpreted in the $L^2$ sense if necessary,Thus a scalar mode bound gives an exact discrete L2 norm bound. Stability on a fixed physical interval requires constants independent of the refining spatial and time meshes:The contraction condition $|G|\le 1$ is sufficient and customary for equations whose exact solutions do not grow. More generally $|G|\le 1+Ck$ suffices, since $(1+Ck{)}^n\le e^{CT}$. At a fixed Courant number, when $G$ is independent of $k$, a mode with modulus strictly greater than one violates the estimate as $n\sim T/k\to \infty$. One must keep this scaling: physical growth such as $e^{\kappa t}$ does not by itself mean finite-time numerical instability. This is finite-time stability versus power boundedness.

For systems, the symbol is an amplification matrix. The requirement is a uniform bound on all its powers, not merely its eigenvalues. A unit-modulus Jordan block $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ has powers $\left(\begin{array}{cc}1& n\\ 0& 1\end{array}\right)$ and is unstable under refinement. A uniformly conditioned eigenbasis or an energy estimate can justify a modal conclusion; eigenvalues alone cannot control a general nonnormal matrix. Multilevel schemes likewise produce an amplification polynomial of a multilevel finite difference scheme, with a uniform root condition and control of starting errors. Repeated unit-modulus roots can give polynomial growth, and roots approaching each other require attention to mesh-uniform constants.

Several elementary examples show how the calculation works. For $u_t+c{u}_x=0$ with $c>0$, forward Euler with centered spatial differences gives $G=1-i\mu \sin \theta$, where $\mu =ck/h$. Since $|G{|}^2=1+{\mu }^2{\sin }^2\theta$, it is unstable at every fixed nonzero Courant number. The upwind finite difference scheme instead gives $G=1-\mu +\mu e^{-i\theta }$ andso its exact contraction range is $0\le \mu \le 1$. The Lax-Wendroff advection scheme has $G=1-i\mu \sin \theta +{\mu }^2(\cos \theta -1)$; expansion gives $|G{|}^2=1-4{\mu }^2(1-{\mu }^2){\sin }^4(\theta /2)$, hence $|\mu |\le 1$. The Courant–Friedrichs–Lewy condition relates these restrictions to the numerical domain of dependence, but is a necessary geometric check rather than a complete stability proof. Implicit schemes can have a global spatial dependence and different Courant ranges, as Question 3 illustrates.

For the heat equation $u_t=\nu u_{xx}$, put $r=\nu k/h^2$. Central spatial differences with forward Euler give $G=1-4r{\sin }^2(\theta /2)$, so $0\le r\le 1/2$. With backward Euler the factor is $[1+4r{\sin }^2(\theta /2){]}^{-1}$, a contraction for every $r\ge 0$. For the Crank-Nicolson diffusion scheme it is $[1-2r{\sin }^2(\theta /2)]/[1+2r{\sin }^2(\theta /2)]$, again a contraction for all $r\ge 0$. At large $r$, however, high-frequency factors approach minus one rather than zero: unconditional stability does not imply strong damping or accurate resolution of every mode.

The connection with convergence is the Lax equivalence theorem for a consistent approximation to a well-posed linear evolution problem. Its forward implication follows directly from the error recursion. If $e^{n+1}=G_h{e}^n+k{\tau }^n$ and $\Vert G_h^j\Vert \le C_T$ for $jk\le T$, thenA convergent initialization and a vanishing normalized local truncation error therefore give convergence. Smooth-data consistency can be extended by density when the approximation and initialization are uniformly bounded. For rough $L^2$ data, cell averages or another L2-compatible initialization of grid data are preferable to undefined point samples. A high formal order does not rescue an unstable amplification operator.

Boundary conditions determine whether Fourier diagonalization is exact. On a periodic grid the update is a circulant matrix, and the finite discrete Fourier transform diagonalizes it at the frequencies $2\pi \ell /N$. Here an interior symbol controls the entire scheme, including the wrapping boundary. On the whole line, the continuous frequency transform serves the same purpose and there is no boundary closure to check.

For a finite interval with homogeneous Dirichlet boundary conditions, centered diffusion is a symmetric tridiagonal problem diagonalized by discrete sine modes. One tests the same diffusion symbol at the admitted sine frequencies, not at arbitrary periodic modes. For homogeneous Neumann boundary conditions, a suitable reflected closure uses discrete cosine modes; the constant mode is present and is neutral for pure diffusion. With the standard reflected endpoint stencil, the natural norm uses half endpoint weights, in which the matrix is self-adjoint. Consequently stable diffusion need not make every solution tend to zero under Neumann conditions. These special orthogonal bases supply an exact substitute for Fourier modes when the operator and closure have the required symmetry.