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The principal strain rates $a,b,c$ are the eigenvalues of the symmetric strain-rate tensor. Incompressibility makes its trace zero, so $a+b+c=0$. The polynomial identitytherefore gives $a^3+b^3+c^3=3abc$. Since the trace of $S^3$ is the sum of the cubes of its eigenvalues, the Betchov relation becomes

A linear operator $Z$ on a Banach space is called a dissipative operator whenOn a Hilbert space, using the inner product linear in its first argument,If $\mathrm{Re}⟨Zf,f⟩\le 0$, this proves the defining inequality. Conversely that inequality implies $2\lambda \mathrm{Re}⟨Zf,f⟩\le \Vert Zf{\Vert }^2$ for every $\lambda >0$; letting $\lambda \to \infty$ gives

For the next spectral assertion, dissipativity must be retained as a hypothesis. Let $Z$ be closed, densely defined and dissipative. For $\alpha >0$, the dissipative bound makes $\alpha I-Z$ injective. Its range is closed: if $(\alpha I-Z)f_n\to h$, the bound makes $f_n$ Cauchy, and then $Z{f}_n=\alpha f_n-(\alpha I-Z)f_n$ converges. Closedness produces a preimage of $h$. If the range is all of $H$, the same bound gives a bounded inverse, so $\alpha$ is in the resolvent set. If the closed range is proper, there is a nonzero vector in its orthogonal complement, and the definition of the adjoint of a densely defined operator givesThis proves that the positive spectrum of a dissipative operator is adjoint point spectrum:Without dissipativity the PDF's assertion is false. For example, multiplication by $x$ on $L^2(0,2)$ is bounded, closed and densely defined. The value $1$ is spectral: normalized functions supported closer and closer to $x=1$ make $\Vert (Z-I)f\Vert$ tend to zero. But $Z^{\ast }=Z$ has no eigenvector for eigenvalue $1$, since a square-integrable function supported on the singleton $\{1\}$ is zero.

For the given maximal-domain operator $Q$, finite-support sequences belong to $D(Q)$, so the domain is dense. Coordinatewise limits also show it is closed: if $f^{(j)}\to f$ and $Q{f}^{(j)}\to h$ in ${\ell }^2$, each defining coordinate equation passes to the limit, giving $Qf=h\in {\ell }^2$ and hence $f\in D(Q)$.

Put $g_n=2^n{f}_n$ and $h=Qf$. For $n\ge 1$,Since ${\sum }_{n\ge 1}{2}^{-n}|h_n|<\infty$ by Cauchy-Schwarz inequality, $g_n$ has a finite limit ${\ell }_f$. Its increments are square summable as well. Direct summation givesThe original inner-product series converges because $Qf,f\in {\ell }^2$. Taking $N\to \infty$ therefore givesThus $Q$ is dissipative, including its full maximal domain; the limiting boundary term has not been discarded.

To decide generation, solve $(\alpha I-Q)f=h$ for arbitrary $h\in {\ell }^2$ and fixed $\alpha >0$. The first coordinate gives $g_0=f_0=h_0/(\alpha +1)$, and the rest giveBecause all denominators are at least one,Hence $g_n$ is bounded, so $f_n=2^{-n}{g}_n$ belongs to ${\ell }^2$. The coordinate equations give $Qf=\alpha f-h\in {\ell }^2$, proving $f\in D(Q)$ and surjectivity of $\alpha I-Q$. Together with density, closedness and dissipativity, the Lumer-Phillips theorem proves the contraction generator from an exponentially weighted difference operator conclusion:

The domain is crucial. The formal transposed matrix suggests a positive-eigenvalue solutionwhich is square summable because the products converge. It is not a true adjoint eigenvector: the admissible test vector $f_n=2^{-n}$ has $Qf=(-1,0,0,\dots )$, so $⟨Qf,u⟩=-\overline{u_0}$, while $\alpha ⟨f,u⟩=\alpha \overline{u_0}{\sum }_n{4}^{-n}{\prod }_{k<n}(1+\alpha 4^{-k})$. These cannot agree for $\alpha >0$ unless $u_0=0$. This exemplifies how boundary terms exclude formal adjoint eigenvectors, and explains why a calculation using only finite-support test vectors would give the wrong generation answer.

Let $K$ denote isotropic permeability of a porous medium, $\mu$ the fluid dynamic viscosity, ${\rho }_0$ the reference fluid mass density, $\alpha >0$ its volumetric thermal-expansion coefficient and $g$ gravitational acceleration. The Boussinesq approximation uses $\rho ={\rho }_0[1-\alpha (T-T_{\ast })]$ only in the gravitational force. With upward coordinate $z$, hydraulic pressure $P$ and Darcy flux $\mathbf{q}$, the equations arewhere $\theta$ is the perturbation to the conductive temperature. To keep porous heat-capacity conventions explicit, let $C_e$ be effective heat capacity per bulk volume, $C_f={\rho }_0{c}_f$ the liquid volumetric heat capacity, $k_e$ effective thermal conductivity, $\sigma =C_e/C_f$, and $\kappa =k_e/C_f$. The thermal equation is $\sigma T_t+\mathbf{q}\cdot \nabla T=\kappa {\nabla }^{2}T$. If the usual unit-capacity convention is adopted, $\sigma =1$ and $\kappa$ is the thermal diffusivity. Thresholds below do not depend on $\sigma$; it only changes temporal rates.

For a conductive state with upward temperature decrease $\Gamma =-{\overline{T}}_z>0$, linearization giveswhere $w=q_z$. The second equation follows by taking the divergence of Darcy law to eliminate $P$ and then differentiating its vertical component. It is important that the Laplacian on its right is horizontal.

In the horizontal layer, both boundary planes prescribe the temperature and are impermeable, so $\theta =w=0$ at $z=0,h$. Use a normal mode proportional to $e^{st+i\mathbf{k}\cdot \mathbf{x}}\sin (n\pi z/h)$, with horizontal wavenumber $k=|\mathbf{k}|$, $n\ge 1$. Writing $b=n\pi /h$, the momentum relation gives $w=B{k}^2\theta /(k^2+b^2)$, and the thermal relation givesA neutral mode therefore requires $B\Gamma /\kappa =(k^2+b^2{)}^2/k^2$. Since $k^2+2b^2+b^4/k^2\ge 4b^2$, the smallest value is $4{\pi }^2/h^2$, attained at $n=1$ and $k=\pi /h$. This derives the onset of convection in a horizontal Darcy layer rather than simply importing its critical number. ThusEvery small disturbance decays for $T_m<T_{m,\mathrm{crit}}$; at equality the critical roll is neutral. Strict decay has this threshold as a supremum, not an attained maximum. The first roll wavelength is $2h$, compatible with the very large horizontal extent. Cooling from above gives the opposite, stable buoyancy feedback.

For the tall square column, define $a$ as its side length and $H\gg a$ as its height. The insulated vertical walls impose ${\partial }_n\theta =0$, while impermeability imposes zero normal Darcy flux. Away from the remote ends, a nearly height-independent mode has hydrostatic pressure adjustment and $w=B\theta$, with zero cross-sectional mean. The transverse temperature modes arewhere $m,n\ge 0$ but $(m,n)=(0,0)$ is excluded: a closed column cannot support a net vertical discharge. Its horizontally uniform density perturbation is balanced by pressure, rather than driving overturning. The rate of a nonconstant mode isThe smallest allowed eigenvalue is ${\pi }^2/a^2$, so the insulated square-column Darcy convection onset isAgain strict decay requires $\Gamma <{\Gamma }_{\mathrm{crit}}$. This is the leading long-column result. End conditions that maintain the basic gradient require the vertical velocity to turn around near the ends and give small finite-height corrections; their exact coefficient cannot be inferred from insulation of the vertical walls alone. For a separated vertical mode of small wavenumber $k_z$, the neutral expression is $B\Gamma /\kappa =(k_{\perp }^2+k_z^2{)}^2/k_{\perp }^2$, which tends to the displayed value as $H/a\to \infty$.

Just above threshold, warm fluid rises along one side and colder fluid sinks along the opposite side, with horizontal return flow at the top and bottom. The two lowest modes, proportional to $\cos (\pi x/a)$ and $\cos (\pi y/a)$, have the same threshold. A square therefore admits either orientation or a superposition, including diagonal overturning. Linear theory predicts the weak columnar circulation and its degeneracy; it does not select a unique nonlinear planform.

For the Volterra integration operator, Cauchy-Schwarz inequality givesIntegrating and reversing the order yieldsThus $J$ is bounded, with $\Vert J\Vert \le 1/\sqrt{2}$. Since $f\in L^2(0,1)\subseteq L^1(0,1)$, $Jf$ is absolutely continuous and $(Jf{)}^{\prime }=f$ almost everywhere. If $Jf=0$ as an $L^2$ element, its continuous representative is zero everywhere, so its derivative is zero almost everywhere. Hence $J$ is one-to-one.

Its range is exactly $\{u\in H^1(0,1):u(0)=0\}$, and $Au=i{u}^{\prime }$. This contains the smooth compactly supported functions in $(0,1)$, so $A$ is a densely defined operator. Closedness can also be seen directly: if $u_n\to u$ and $A{u}_n\to v$ in $L^2$, then $u_n=J(-iA{u}_n)\to J(-iv)$, giving $u=J(-iv)\in D(A)$ and $Au=v$.

For every $\lambda \in \mathbb{C}$,If $\lambda \ne 0$, the allowed fact $\sigma (J)=\{0\}$ makes $J-(i/\lambda )I$ invertible. At $\lambda =0$, $\lambda J-iI=-iI$ is invertible too. Thus the bounded operatorhas range in $D(A)$ and is a two-sided inverse of $\lambda I-A$. Equivalently, solving the initial-value differential equation givesThis is bounded for every fixed $\lambda$ on the finite interval. Therefore Volterra inverse differentiation has empty spectrum:The familiar nonempty-spectrum theorem for bounded operators does not apply to this unbounded operator.

The subspace $H_0$ is the kernel of the bounded functional $f\mapsto Jf(1)={\int }_0^{1}f$, so it is closed. Its image under $J$ isthe zero-trace Sobolev space. If $u_n\in D(A_0)$, $u_n\to u$ and $A_0{u}_n\to v$, then $-i{A}_0{u}_n\in H_0$ converges to $-iv\in H_0$, and $u=J(-iv)\in D(A_0)$. Thus $A_0$ is closed. It is densely defined because it also contains the smooth compactly supported functions.

For $u,v\in D(A_0)$, integration by parts and their zero endpoint traces giveHence $A_0$ is closed and symmetric.

For any $v\in C^1([0,1])$, the same formula gives $⟨A_{0}u,v⟩=⟨u,i{v}^{\prime }⟩$, with no endpoint term because $u$ vanishes there. Consequently $v\in D(A_0^{\ast })$ and $A_0^{\ast }v=i{v}^{\prime }$, proving the requested inclusion. In fact, testing against compactly supported smooth $u$ shows that every element of $D(A_0^{\ast })$ has weak derivative in $L^2$; conversely the integration-by-parts formula applies to every $v\in H^1(0,1)$. Thuswith no boundary restrictions.

For any $\lambda \in \mathbb{C}$, the nonzero function $v_{\lambda }(x)=e^{-i\lambda x}$ belongs to this domain and satisfies $i{v}_{\lambda }^{\prime }=\lambda v_{\lambda }$. Conversely this first-order equation has only scalar multiples of that exponential. ThereforeFor any $\lambda$, the adjoint eigenfunction with eigenvalue $\overline{\lambda }$ annihilates $\mathrm{Ran}(\lambda I-A_0)$, so this range is not all of $H$. Explicitly the resolvent equation with zero initial value has the solution $R_{\lambda }h$ above, and its endpoint value is zero exactly whenThus the range is a proper closed hyperplane for every $\lambda$. This proves that the two-endpoint symmetric derivative has whole complex spectrum:There are no eigenvalues of $A_0$ itself: the eigenfunction equation together with $u(0)=0$ forces the exponential coefficient to vanish. The operator is symmetric but not self-adjoint, since its adjoint has a strictly larger domain; symmetry alone does not force a real spectrum.

Since the derivative of $\tanh (\beta x)$ at zero is $\beta$, the Jacobian matrix of this directed rate network isWith $a=1+\tau \lambda$, its characteristic determinant isThus the eigenvalues satisfyThis establishes three-neuron ring stability from the loop product. The ring is directed, so the symmetric-weight Hopfield network energy argument does not apply. Assume the physical relaxation time $\tau >0$.

An even number of negative weights gives $\Gamma >0$ when all three connections are nonzero. Put $r={\Gamma }^{1/3}$. The three cube roots giveThe complex pair always has negative real part, and the real eigenvalue is negative exactly when $r<1$. Therefore strict linear asymptotic stability requires $\Gamma <1$. The origin is unstable when $\Gamma >1$. A zero connection gives $\Gamma =0$ and all eigenvalues $-1/\tau$, also stable, though the matrix need not be diagonalizable.

At $\Gamma =1$, linearization has a simple zero eigenvalue, so a strict negative-real-part test alone does not settle nonlinear stability. For completeness, take a nonzero real eigenvector $c$ of the weighted loop matrix $M=\beta W$, satisfying $Mc=c$, and set $x_i=c_i{u}_i$. The local equations areThe center direction has all $u_i=a$. Averaging the three equations on the center manifold gives $\tau \dot{a}=-({\beta }^2{\sum }_i{c}_i^2/9)a^3+O(a^5)$. Its leading coefficient is negative, while the other modes decay exponentially. Hence the exact boundary origin is locally asymptotically stable for the specified hyperbolic-tangent nonlinearity, with algebraic rather than exponential decay. This qualification distinguishes actual nonlinear stability from the usual strict Jacobian criterion.

Put $S=u^2+v^2$. At a nonzero equilibrium point the two equations form the homogeneous linear systemIts determinant must vanish, giving $S^2/2-(\mu +\sigma )S+{\mu }^2+{\sigma }^2-1=0$. ThusEach positive real polynomial root gives an antipodal equilibrium point pair; a negative polynomial root does not describe a real equilibrium point. A direction is specified by $\sin 2\theta =S/2-\mu$, $\cos 2\theta =S/2-\sigma$, where $u=\sqrt{S}\cos \theta$, $v=\sqrt{S}\sin \theta$. This is the polar form of the cubic confinement model. The origin is always an equilibrium point.

The origin's eigenvalues are ${\lambda }_{\pm }=\mu \pm \sqrt{1-{\sigma }^2}$. A simple zero eigenvalue occurs on ${\mu }^2+{\sigma }^2=1$, except at $\mu =0$. The symmetry of an odd function $(u,v)\mapsto (-u,-v)$ makes the generic steady bifurcation a pitchfork bifurcation. At a point $(\sigma ,{\mu }_0)$ on this circle with ${\mu }_0\ne 0$, vary $\mu ={\mu }_0+\delta$. Expanding the radius equation gives $S\sim 2{\mu }_0\delta /({\mu }_0+\sigma )$. In a signed, unit-eigenvector centre manifold coordinate, the cubic coefficient therefore has sign $-({\mu }_0+\sigma )/(2{\mu }_0)$.

On the lower semicircle ${\mu }_0=-\sqrt{1-{\sigma }^2}$ the transverse eigenvalue is negative: the pitchfork bifurcation is supercritical for $\sigma <1/\sqrt{2}$ and subcritical for $\sigma >1/\sqrt{2}$. On the upper semicircle it is supercritical for $\sigma >-1/\sqrt{2}$ and subcritical for $\sigma <-1/\sqrt{2}$ in the centre manifold direction, but its transverse eigenvalue is positive, so the bifurcating equilibrium points are not fully attracting. At $(\sigma ,{\mu }_0)=(\pm 1/\sqrt{2},\mp 1/\sqrt{2})$ the cubic vanishes. The radius equation instead gives $\delta =-S^2/(4{\mu }_0)+o(S^2)$, a nonzero quintic degeneracy with amplitude proportional to $|\delta {|}^{1/4}$: these are degenerate pitchforks in the cubic confinement model.

A Hopf bifurcation occurs at $\mu =0$, $|\sigma |>1$, with frequency $\sqrt{{\sigma }^2-1}$. To determine its direction, first note that the divergence is $2\mu -2S$. The Bendixson-Dulac criterion excludes every nonconstant periodic orbit for $\mu \le 0$, since this divergence is negative except possibly at the origin. We can also calculate a nonzero saturation coefficient. Set $a=|\sigma +1|$, $b=|\sigma -1|$ and $V=a{u}^2+b{v}^2$. At $\mu =0$ the linear orbits are ellipses of constant $V$. Direct differentiation givesOn a linear ellipse, write $u=\sqrt{V/a}\cos \varphi$, $v=\sqrt{V/b}\sin \varphi$. The angular speed is constant; the averages of $u^{3}v$ and $u{v}^3$ vanish, while $⟨S⟩=V|\sigma |/({\sigma }^2-1)$. Averaging the cubic terms, or removing their oscillatory parts by a periodic change of coordinates in its normal form, gives for $\varrho =\sqrt{V}$The cubic coefficient is strictly negative. Hence a small attracting limit cycle exists for $\mu >0$, andFinally $(\sigma ,\mu )=(\pm 1,0)$ have a nonzero nilpotent matrix as their linearization with two zero eigenvalues. They are reflection-symmetric double-zero bifurcations where the Hopf bifurcation and steady thresholds meet; an ordinary nonzero-frequency Hopf bifurcation calculation does not apply there. These, the two degenerate pitchfork bifurcation points, and the generic pitchfork bifurcation/Hopf bifurcation loci above exhaust the origin's local loss-of-hyperbolicity possibilities.

The trace and determinant of the origin's linearization are $2\mu$ and ${\mu }^2+{\sigma }^2-1$. Thus its open region of linear asymptotic stability isFor $|\sigma |<1$ this means below the lower semicircle; for $|\sigma |\ge 1$ all negative $\mu$ lie in the open stable region. The origin is a saddle equilibrium inside the unit circle and a source outside it with positive $\mu$.

For nonzero equilibrium points, put $\Delta =2-(\mu -\sigma {)}^2$. Existence requires $\Delta \ge 0$ and $\mu +\sigma +\sqrt{\Delta }>0$. An equivalent region isInside the circle exactly one radius polynomial root is positive, giving two equilibrium points. Outside it, with $\mu +\sigma >0$ and $\Delta >0$, both polynomial roots are positive, giving four. On $\Delta =0$ with $\mu +\sigma >0$, the two radii coalesce in a saddle-node bifurcation at each antipodal point. On the circle a zero radius is the origin and is excluded from the nontrivial count. These distinctions are shown in the parameter sketch.

At $(\sigma ,\mu )=(1/\sqrt{2},-1/\sqrt{2})$ the linearization has one simple zero and one negative eigenvalue, while the cubic centre manifold coefficient vanishes. The quintic coefficient is negative. It is a codimension-two bifurcation with a reflection-symmetric degenerate pitchfork bifurcation: a wedge of coexistence between an attracting origin and an attracting antipodal pair opens between a subcritical pitchfork bifurcation and the nearby saddle-node bifurcation curve. The intervening smaller-radius pair consists of saddles. Indeed at a nonzero equilibrium point, direct differentiation of the polar coordinates equations givesThe smaller-radius branch has negative determinant, while in the coexistence wedge the larger-radius branch has positive determinant and negative trace.

At critical boundaries the linear stability analysis test alone is inconclusive. The origin is nonlinearly attracting at a supercritical lower pitchfork bifurcation, at its stabilizing quintic endpoint, and at the supercritical Hopf bifurcation threshold, although decay is no longer exponential. It is unstable at the subcritical lower pitchfork bifurcation. The double-zero point $(\sigma ,\mu )=(-1,0)$ is also attracting: the positive function $W=v^2+u^4/8$ hasnear the origin away from zero. Here $|u{v}^3|\le (u^2{v}^2+v^4)/2$ and $|u^{3}v|\le (u^4+u^2{v}^2)/2$ prove the bound. The other double-zero point $(1,0)$ has an unstable quartic potential, as the blow-up below shows. The shading records the open linear stability analysis regions; these boundary remarks distinguish nonlinear from exponential stability.

At $(\sigma ,\mu )=(1,0)$ the origin's matrix is $\left(\begin{array}{cc}0& 0\\ 2& 0\end{array}\right)$: it has a double zero eigenvalue and one eigenvector. Trace and determinant vary independently with $\mu$ and $\sigma$ there, so two parameters are needed. The symmetry of an odd function removes quadratic terms, giving a reflection-symmetric cubic double-zero normal form. The quadratic nonlinearity required for a generic Bogdanov–Takens bifurcation vanishes.

Use a prime for differentiation with respect to the rescaled time. Direct substitution, with all rescaled variables written without tildes, givesAt $\epsilon =0$, the equations are $u^{\prime }=-sv+v^3/2$, $v^{\prime }=2u$. Fordifferentiation gives $H^{\prime }=2u(-sv+v^3/2)+(sv-v^3/2)2u=0$. This is the Hamiltonian blow-up of a reflection-symmetric double-zero point.

For $s>0$, the equilibrium points are a centre equilibrium at $(0,0)$ and saddles at $(0,\pm \sqrt{2s})$. The centre equilibrium has eigenvalues $\pm i\sqrt{2s}$; at either saddle equilibrium they are $\pm 2\sqrt{s}$. Levels $0<H<s^2/2$ contain closed inner ovals around the centre equilibrium. The saddle equilibrium level hasThese two branches form a heteroclinic cycle connecting the saddles. The branch with $u>0$ runs from the lower saddle equilibrium to the upper one because $v^{\prime }=2u>0$; the other runs back. Outside the potential well, trajectories are open. The right panel of the preceding figure uses the coordinates $(v,u)$ to display the well and these directed connections.

Write $k=\sqrt{\mu }$. The equilibrium points are a stable node $(-k,0)$ and a saddle equilibrium $(k,0)$, whose unstable and stable eigenvalues are $2k$ and $-\lambda$. The saddle equilibrium's local stable manifold is $x=k$. When the right-going unstable separatrix returns at $(\nu ,h)$ with $\nu =k$, it lands exactly on that stable manifold, producing a homoclinic orbit of the saddle. Crossing this condition changes whether the returning trajectory is captured by the node or lies on the outgoing side of the saddle equilibrium.

Use an incoming section $y=h$ and outgoing section $x=h$. For entry $(x,h)$ with $k<x<h$, integrate the local flow:so the outgoing height isThe global excursion is smooth and transverse. Its return coordinate has expansion $x_{\mathrm{next}}=\nu +cY+O(Y^2)$, with $c>0$ in these section orientations. Thus the local passage map near a dissipative saddle-node givesFor small $\mu >0$, $\lambda >2k$, hence $P(k^{+})=\nu$ and $P^{\prime }(k^{+})=0$. If $\nu -k>0$ is small, the graph starts above the diagonal and, with slope less than one nearby, crosses it close to $x=\nu$. More explicitly, $P(x)-x=(\nu -k)-(x-k)+O((x-k{)}^{\beta })$, giving a unique small fixed point with Floquet multiplier $0<P^{\prime }<1$. It corresponds to an attracting periodic orbit. For $\nu <k$ no such local crossing exists on $x>k$; the unstable branch returns to the node's side instead. The Poincaré return map panel shows this geometric fixed-point test.

Linearizing about the orbit gives $\dot{\eta }=A(t)\eta$, where $A(t)=Df(X(t))$. The fundamental matrix therefore satisfiesThe Floquet multipliers are the eigenvalues of the monodromy matrix $\Phi (T)$. Differentiating the orbit equation shows that its tangent $\dot{X}(t)$ solves the variational equation. Since $\dot{X}(T)=\dot{X}(0)\ne 0$, one multiplier is one.

For an invertible fundamental matrix, differentiating the determinant gives $(\det \Phi {)}^{\prime }=\mathrm{tr}({\Phi }^{-1}\dot{\Phi })\det \Phi =\mathrm{tr}A\det \Phi$. Integration from the identity yields $\det \Phi (T)=\exp ({\int }_0^T\mathrm{tr}A(t)dt)$. The determinant is the product of the two multipliers, so the other is

Positive definiteness of $A$ implies $D>0$. Let $B=D^{-1/2}A{D}^{-1/2}$, a symmetric positive definite matrix. The additional hypothesis says $2I-B=D^{-1/2}(2D-A)D^{-1/2}>0$, so all eigenvalues $\lambda$ of $B$ satisfy $0<\lambda <2$. The ordinary Jacobi error matrix $I-D^{-1}A$ is similar to $I-B$ and therefore has eigenvalues $1-\lambda \in (-1,1)$. Its spectral radius is less than one. The convergence criterion for stationary linear iterations consequently gives convergence from every starting vector.

For an $n\times n$ matrix here, $D=I$. The sine vectors $v_j(i)=\sin (ij\pi /(n+1))$ give eigenvaluesThus every eigenvalue lies strictly between $1/2$ and $3/2$. For $0<\omega \le 4/3$, $0<\omega {\lambda }_j<2$, so $|1-\omega {\lambda }_j|<1$, including the upper endpoint $\omega =4/3$. Hence the iteration converges for the entire stated range.

The extreme eigenvalues are $1-a$ and $1+a$, with $a=\frac{1}{2}\cos (\pi /(n+1))$. The spectral radius is $\max (|1-\omega (1-a)|,|1-\omega (1+a)|)=|1-\omega |+a\omega$ for $\omega >0$. It decreases up to one and increases after one, since $0\le a<1$. ThereforeFor $n=1$ the same formula gives zero spectral radius and exact solution in one ordinary Jacobi step.