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Set $T=ϵt$ and use the method of multiple scales:Since ${\omega }^2=1/4+kϵ+O({ϵ}^2)$, the first correction satisfiesThe resonant cosine and sine coefficients must vanish to avoid secular terms. Using the product-to-sum formulas givesThe remaining forcing has frequency $3/2$ and permits the bounded particular correction $y_1=(a/4)\cos (3t/2)+(b/4)\sin (3t/2)$. The two slow amplitudes carry the two arbitrary constants of the original equation.

For an explicit general leading solution, putThe matrix exponential givesThe error is uniform on bounded intervals of $T$, for bounded initial data. When $\lambda$ is imaginary this expression uses ordinary sine and cosine; at $\lambda =0$ its continuous limit is $I+TM$.

The primary instability tongue of a weak Mathieu oscillator has a positive slow eigenvalue precisely whenThese are the leading edges, rather than an exact finite-$ϵ$ interval. At the leading endpoints the amplitude matrix is a nilpotent matrix, so the leading solution can grow linearly in $T$ and does not decide the exact exponential instability. For completeness, set $\omega =1/2+\beta ϵ+\gamma {ϵ}^2$. The periodic or antiperiodic edge calculation uses a pure cosine when $\beta =-1/2$ and a pure sine when $\beta =1/2$, with the third-harmonic correction just found. The next resonant coefficient is $-(\gamma +3/8)y_0$, soThus the exact growing interval is between these shifted edges. In particular, for positive small $ϵ$ the fixed leading value $k=-1/2$ lies just inside it, whereas $k=1/2$ lies just outside. Interior detunings bounded away from the edges are already resolved by the leading method of multiple scales.

For the state vector $q=(h,u,g^{\prime }{)}^T$, the primitive equations are $q_t+\mathsf{A}{q}_x=S$, withThe characteristic polynomial isConsequently the three characteristic speeds areFor $h>0$ and $g^{\prime }>0$, the hyperbolic system has three distinct real eigenvalues. The long-wave speed relative to the current is $c$, while $u$ transports a buoyancy/contact disturbance.

A left eigenvector for the material characteristic is $(0,0,1)$, givingFor $u\pm c$, left eigenvectors can be chosen asMultiplying $q_t+\mathsf{A}{q}_x=S$ by these rows gives the characteristic compatibility for an entraining triangular-channel current. With all differentials evaluated along the indicated characteristic,The source follows directly from ${\ell }_{\pm }S=-2u{w}_e/h\pm 2c{w}_e/h\mp 2g^{\prime }{w}_e/(3c)$ and $g^{\prime }/c=2c/h$.

Since $4dc=(2c/h)dh+(h/c)d{g}^{\prime }$, an equivalent form isCombining it with the material buoyancy equation also givesOnly when entrainment vanishes and $g^{\prime }$ is spatially constant do $u\pm 4c$ become constant Riemann invariants on these characteristic curves. Dropping the buoyancy differential in the general problem would lose one of the required transport couplings. The characteristic formulas with $1/c$ apply to the nondegenerate region $g^{\prime }h>0$; dry or neutrally buoyant limits require their conservative form instead.

For a linear system write $A=D+R$, where $D$ is the diagonal. The Jacobi method iteratesHere $D=(10/3)I$, and $B$ has nonnegative off-diagonal entries with row sums at most one. Rows adjacent to the boundary have sum strictly less than one; those at purely interior stencil vertices have sum one. Thus $\Vert B{\Vert }_{\infty }\le 1$.

To prove strict spectral radius below one, suppose $Bv=\zeta v$ with $|\zeta |=1$, and choose an index with $|v_i|=\Vert v{\Vert }_{\infty }>0$. Inequality forces that row sum to equal one and every linked neighbor to attain the same maximum modulus. Propagate this along the connected axial grid to a boundary-adjacent row, whose row sum is strictly less than one, a contradiction. All eigenvalues therefore have modulus strictly below one. Hence $B^k\to 0$ and the Jacobi errors tend to zero from every initial vector. The unique solution exists by positive definiteness, so $\boxed{\text{Jacobi converges for this discretization in any ordering}.}$

For general $k$, the same energy gives $\dot{E}=-k{y}^2$. For $k=0.1$ the eigenvalues at the origin are $-0.05\pm i\sqrt{3.99}/2$, so it is a stable focus. At $(1,0)$ they are $(-0.1\pm \sqrt{4.01})/2$, so it remains a saddle equilibrium. The vector field has constant divergence $-0.1$: there are no periodic orbits, by the planar divergence criterion. Nor can the former homoclinic orbit survive, since $E$ strictly decreases along every nonconstant orbit except at isolated crossings of $y=0$.

The stable manifold of the saddle separates escape to positive $x$ from convergence to the origin. Its two branches approach the saddle in forward time with tangent slope $(-0.1-\sqrt{4.01})/2$. The left unstable branch spirals into the origin, whereas the right unstable branch escapes. The shaded component on the origin's side of the stable manifold in the right panel is the basin of attraction. The boundary itself is excluded: its trajectories converge to the saddle. This boundary is the damped stable manifold, rather than the conservative energy contour. energy decrease prevents bounded recurrent motion, and every bounded orbit off the saddle's stable manifold consequently converges to the origin. The plotted manifold was obtained by integrating backwards from the saddle's stable eigendirection.

For $k=0$, the first integralhas $\dot{E}=0$. The equilibrium points are $(0,0)$ and $(1,0)$. The linearization has eigenvalues $\pm i$ at the origin and $\pm 1$ at $(1,0)$. The origin is a center equilibrium: the sufficiently small positive level sets of $E$ are closed curves, since its Hessian matrix there is positive-definite. The other equilibrium is a saddle equilibrium.

At the saddle, $E=1/6$, and the separatrix satisfiesIts two branches with $-1/2\le x\le 1$ form a homoclinic orbit through the saddle; its remaining branches extend to $x>1$. Inside the loop, the closed trajectories run clockwise: above the horizontal axis $\dot{x}=y>0$. The saddle directions have slopes $+1$ and $-1$. Outside the loop there are unbounded trajectories. The left panel shows the closed orbits and all saddle branches.

At a stationary bifurcation, a real eigenvalue of the equilibrium linearization crosses zero; the critical motion has zero frequency. At a Hopf bifurcation, a conjugate pair crosses the imaginary axis at $\pm i\omega$ with $\omega >0$, permitting a nearby limit cycle.

The Hopf normal form has a small cycle when $R^2=\mu /a+O({\mu }^2)>0$. Its radial multiplier exponent is ${\partial }_r\dot{r}{|}_R=-2\mu +O({\mu }^2)$, while the origin's radial exponent is $\mu$. Thus: for $\mu <0<a$, there is an attracting spiral and no small cycle; for $0<\mu ,a$, a repelling spiral is surrounded by an attracting cycle; for $\mu ,a<0$, an attracting spiral is surrounded by a repelling cycle; for $a<0<\mu$, there is a repelling spiral and no small cycle. Outside the repelling cycle in the third case the local radial flow points outward; a cubic normal form makes no global prediction about where it goes.

The subcritical bifurcation has $a<0$ and the unstable radius branch $R=\sqrt{\mu /a}+O(|\mu {|}^{3/2})$ on $\mu <0$, meeting the origin at $\mu =0$. On either existing branch, the angular speed is $\omega +(c-b/a)\mu +O({\mu }^2)$, so

For a flow, a hyperbolic fixed point has no eigenvalue of its linearization with zero real part; a nonhyperbolic fixed point has at least one such eigenvalue. At $(\mu ,0)$ the eigenvalues are $-\mu ,1-\mu$, while at $(0,\pm 1)$ they are $\mu +1,-2$. Thus nonhyperbolicity occurs at the two stated parameters. The following centre manifold reductions show the associated actual bifurcations.

Near $(1,0)$ put $X=x-\mu$, $Y=y$, $\nu =\mu -1$. The transformed equations areAt $\nu =0$, the non-extended stable subspace is the $X$ axis and the centre subspace is the $Y$ axis. In the extended centre manifold, write $X=h(Y,\nu )$, with $h(0,\nu )=0$. The symmetry $Y\mapsto -Y$ allows an even graph. At quadratic order its invariance equation has left side of degree at least three and right side $-h+Y^2$, so $h=Y^2+O(\nu Y^2,Y^4)$. Substitution givesThis is a supercritical pitchfork bifurcation in parameter $-\nu =1-\mu$. The central branch is stable for $\mu >1$ and unstable for $\mu <1$ near one; the two nonzero branches for $\mu <1$ are stable. At the bifurcation itself the cubic term attracts on the centre manifold.

Near $(0,1)$ put $X=x$, $Y=y-1$, $\nu =\mu +1$. ThenAt $\nu =0$, the stable subspace is $X=0$ and the centre line is $Y=-X/2$. Thus the leading extended centre graph is $Y=-X/2+O(X^2,\nu X)$, and the reduced flow isThis is a transcritical bifurcation: $X=0$ is stable for $\nu <0$, while the branch $X=\nu /2+\cdots$ is stable for $\nu >0$; the opposite branches are unstable. At the critical point the quadratic term is only one-sided attracting.

The exact additional branch in $y>0$ isIt meets $(\mu ,0)$ at $\mu =1$ and $(0,1)$ at $\mu =-1$, displaying the connecting pitchfork and transcritical branches in a quadratic-product flow. Its Jacobian has determinant $1-{\mu }^2$ and trace $(\mu -3)/2$. Hence it is stable for $-1<\mu <1$ and a saddle for $\mu <-1$. The negative-$y$ companion is needed for the pitchfork sketch near $(1,0)$; no separate analysis of $(0,-1)$ is required.

A periodic lattice Hamiltonian commutes with lattice translations. The Bloch theorem therefore classifies its states by characters of the translation group:where $u$ is lattice periodic. For each crystal momentum $\mathbf{k}$, solving on a primitive cell with the corresponding quasiperiodic boundary condition gives discrete eigenvalues indexed by $n$. Their continuous dependence on $\mathbf{k}$ produces the energy bands $E_n(\mathbf{k})$ as $\mathbf{k}$ ranges through the Brillouin zone. Bragg mixing opens gaps at suitable degeneracies, although bands can overlap in three dimensions and gaps are not guaranteed between every pair.

The quantum numbers are band index, wavevector modulo reciprocal lattice vectors, and spin where spin is conserved. In a finite crystal the allowed wavevectors are discrete; in the thermodynamic limit they become dense and the band description emerges. Spin-orbit coupling can replace a separately conserved spin label by an appropriate internal band label.

Dilute randomly placed impurities break exact translation symmetry, so crystal momentum is no longer an exact quantum number. Most extended states retain approximately the host band structure but acquire impurity scattering and disorder broadening. An impurity potential can also produce impurity levels in a crystal, often in a band gap: levels associated with a donor impurity lie near the conduction-band edge and can supply electrons, while those associated with an acceptor impurity lie near the valence-band edge and can create holes. The impurity's species and potential determine whether such levels exist and how deep they are; not every impurity necessarily makes a gap state. At low density these levels are nearly isolated, while overlap at increasing density can broaden them into an impurity band. Added or removed carriers shift the chemical potential as well.

Since $A$ is diagonalizable over the complex numbers, write $x^{(0)}={\sum }_i{a}_i{v}_i$. Provided iteration does not encounter zero, its direction is that ofWith one strictly dominant eigenvalue and $a_n\ne 0$, all other contributions are smaller by factors $(|{\lambda }_i|/|{\lambda }_n|{)}^k$. The dominant eigenvalue is real, since nonreal eigenvalues of a real matrix occur in conjugate pairs of equal modulus. The normalized power method therefore converges to a signed unit eigenvector if ${\lambda }_n>0$, and alternates between opposite unit eigenvectors if ${\lambda }_n<0$. A quotient such as $(x^{(k)}{)}^{T}A{x}^{(k)}$ approaches ${\lambda }_n$ in either case. If $a_n=0$, replace it by the largest active modulus; if $A{x}^{(k)}=0$, the prescribed iteration is undefined.

With two dominant eigenvalues of equal modulus and both components active, the iterates approach their real invariant plane rather than one direction. Distinct real equal-modulus eigenvalues are $\rho ,-\rho$ and produce a limiting two-cycle. A conjugate pair $\rho e^{\pm i\theta }$ produces a rotation in suitable coordinates on that plane: the normalized directions are periodic when the angle is rational relative to $2\pi$ and otherwise generally keep rotating without convergence. If the two eigenvalues are equal and real, the initial projection is itself one eigenvector in their repeated eigenspace, and converges or alternates as in the single-value case; one orbit cannot recover every independent eigenvector of that eigenspace.

For distinct dominant values, retain unnormalized consecutive vectors $y,Ay,A^{2}y$ in the nearly invariant plane and solveThe roots of ${\lambda }^2-s\lambda +p$ estimate the two eigenvalues. Corresponding eigenvectors are $Ay-{\lambda }_{2}y$ and $Ay-{\lambda }_{1}y$, since applying $A$ and using the recurrence multiplies them by ${\lambda }_1$ and ${\lambda }_2$. This is dominant eigenpair extraction from a two-step Krylov recurrence. Equivalent extraction uses the $2\times 2$ matrix of $A$ on the computed plane. It needs two independent iterates and nonzero projections onto both distinct eigenvectors.

The given vectors satisfyand $y_k,y_{k+1}$ are independent, so their span is invariant under $A$. On that plane the characteristic polynomial is ${\lambda }^2-2\lambda +2$. ThereforeIndeed $A{v}_{\pm }=y_{k+2}-(1\mp i)y_{k+1}=(1\pm i)v_{\pm }$. No third eigenvalue is determined by these data.

The displayed $x_k=(1,1,1)$ has norm $\sqrt{3}$, although a large-$k$ normalized iterate should have norm one. This harmless source normalization is resolved by dividing all three displayed vectors by $\sqrt{3}$; the recurrence, eigenvalues and eigenvector directions are unchanged.

The shifted matrix $N=D-\lambda I$ is still a normal matrix, sinceand $N{N}^{†}$ has the same expansion except that its first term is $D{D}^{†}$, which equals $D^{†}D$. For the given eigenvector, $Ne=0$. Part (b), applied to $N$, therefore gives $N^{†}e=0$. Since $N^{†}=D^{†}-\overline{\lambda }I$,The corresponding eigenvalue of the Hermitian conjugate is the complex conjugate of the original eigenvalue.