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Here $D=I$, so the Jacobi iteration matrix isThe all-ones vector is an eigenvector with eigenvalue $-2\alpha$. On its two-dimensional orthogonal complement, the coordinates sum to zero and $Hv=\alpha v$, so the other eigenvalue is $\alpha$ with multiplicity two. ThereforeBy Jacobi convergence for a three-by-three equicorrelation matrix, convergence occurs exactly forAt either endpoint the spectral radius is one, so convergence for arbitrary initial data fails.

Choose a matrix splitting $A=M-N$ with $M$ invertible. Then $Ax=b$ is equivalent toand the associated stationary iterative method for a linear system isFor the Jacobi method, write $A=D+L+U$ with $D$ diagonal and $L,U$ strictly lower and upper triangular, and takeThusThe iteration converges for every initial vector exactly when the spectral radius of its iteration matrix satisfies

The one-dimensional Isentropic Euler equations areFor the polytropic equation of state,The continuity equation therefore implieswhile the momentum equation becomesAdding and subtracting these equations givesThus the Riemann invariantsare constant along the characteristic curves

Away from the rim's $O(h)$ edge region, the geometry and uniform injection have no radial scale other than the factor required by continuity. It is therefore consistent to take $w=w(z)$. Regularity at $r=0$ then lets the continuity equation integrate toThe no-slip boundary condition and prescribed normal velocities areSince $p$ is independent of $z$, the radial equation implies that $w^{\prime \prime \prime }$ is constant. The four boundary conditions giveand henceUsing $w^{\prime \prime \prime }=12V/h^3$ in $p_r=\mu u_{zz}$ givesTaking the pressure at the rim to be atmospheric, $p(R)=p_0$, produces the porous-plate lubrication cushion pressureThe upward pressure force balances the disc's weight:Since $\epsilon =h/R$,

Let $u(r,z)$ and $w(r,z)$ be the radial and vertical velocity components. For steady incompressible axisymmetric flow without swirl, mass conservation and the relevant cylindrical components of the Navier-Stokes equation are

The vertical scale is $w\sim V$. Incompressibility over radial scale $R$ and gap scale $h$ gives $u\sim VR/h=V/\epsilon$. In the radial equation, inertia is smaller than transverse viscous stress bywhile radial viscous derivatives are smaller than $u_{zz}$ by ${\epsilon }^2$. The pressure scale is $\mu V{R}^2/h^3$. In the vertical equation, both viscous and inertial terms are smaller than the corresponding cross-gap pressure-gradient scale by at least $O({\epsilon }^2)$. Thus the leading lubrication theory equations are

The Killing equation gives ${\nabla }_{\mu }{K}^{\mu }=0$ and ${\nabla }_{[\mu }{K}_{\lambda ]}={\nabla }_{\mu }{K}_{\lambda }$. Commuting derivatives once and contracting the Ricci identity givesTherefore, using the contracted Bianchi identity and symmetry of $R_{\mu \beta }$,Since ${\nabla }_{[\mu }{K}_{\lambda ]}$ is antisymmetric, swapping the two contracted index names shows that the left side equals its derivative-antisymmetrized form:Now set $A_{\mu \lambda }={\nabla }_{[\mu }{K}_{\lambda ]}$. The right side is the contracted commutator $[{\nabla }^{\mu },{\nabla }^{\lambda }]A_{\mu \lambda }$. By the curvature commutator on a covariant tensor, it is a sum of contractions of the symmetric Ricci tensor with the antisymmetric tensor $A$, and hence vanishes. Thusso the Killing vector preserves scalar curvature.

Lower the free index of $K$. The Killing equation makes the last two indices of ${\nabla }_{\nu }{\nabla }_{\mu }{K}_{\alpha }$ antisymmetric after the appropriate derivative is moved. Combining the three resulting Ricci identities cyclically givesApplying the covector form of the Ricci identity and then the first Bianchi identity reduces this toRaising $\alpha$ proves the second covariant derivative of a Killing vector identityContracting $\mu$ and $\nu$ and using the definition and symmetries of the Ricci tensor gives

Expanding the two covariant derivatives of a vector, the second partial derivatives cancel in the commutator. The remaining derivatives and products of Christoffel symbols combine into the coordinate definition of the Riemann curvature tensor, giving the Ricci identityFor a type-$(2,0)$ tensor, the connection acts on both upper indices, so the curvature commutator on a contravariant tensor is

Because the Levi-Civita connection is torsion-free, its Christoffel symbols satisfy ${\Gamma }_{\rho \alpha }^{\mu }={\Gamma }_{\alpha \rho }^{\mu }$. Substitute the stated coordinate expression for each term in the cyclic sumThe partial-derivative terms cancel pairwise by the lower-index symmetry of $\Gamma$, and the quadratic terms cancel after relabelling their dummy indices. Thereforewhich is the first Bianchi identity.

At fixed volume, entropy from heat capacity obeys $dS=C(T)dT/T$. The third law of thermodynamics fixes $S(0)=0$ when the ground state is nondegenerate, so below the transitionAbove the transition,For a transition with no latent heat, entropy is continuous, and comparison with $S=\beta \log T+\gamma$ gives the directly testable relationThus calorimetric measurements of $C(T)$ and $T_c$ verify $\gamma$. More generally, if a measurable latent heat $L$ is present, $S(T_c^{+})-S(T_c^{-})=L/T_c$, and the formula becomes

Macroscopically, an adiabatic process exchanges no heat with its surroundings: $\delta Q=0$. A reversible adiabatic process is therefore isentropic, since $dS=\delta Q_{\mathrm{rev}}/T=0$. Microscopically, the system is thermally isolated, so changes in its energy arise from work done by changing external parameters rather than random energy exchange with a heat bath; in a sufficiently slow change, the populations of the corresponding microscopic states are carried along without producing entropy.

For repeated operation the working substance must execute a thermodynamic cycle. Reversible isothermal contact at $T_2$ absorbs heat $Q_2$, a reversible adiabat changes the temperature to $T_1$, reversible isothermal contact at $T_1$ rejects heat $Q_1>0$, and a second adiabat returns to the initial state. This is the Carnot cycle. Reversibility and zero net entropy change giveThe thermal efficiency is the net work output divided by the absorbed heat:ConsequentlyThe Carnot theorem shows that any irreversible cycle has smaller efficiency, so the two isotherms joined by two adiabats are optimal.

Keep $\mathbf{A}=(0,Bx,0)$ and choose the scalar potential $\varphi =-Ex$. ThenFor $p_z=0$, completing the square givesHence the spectrum on ${\mathbb{R}}^3$ is

For a translationally invariant dispersion, the group velocity is $v_y={\hslash }^{-1}\partial E_n/\partial k$. Thereforein agreement with the electric-cross-magnetic-field drift $\mathbf{E}\times \mathbf{B}/B^2$.

In the rectangle, $k=2\pi r/L_y$ as before. The energy now changes with $r$, so the guiding-centre degeneracy is lifted, while the number of states in each tilted band remains approximately $|qB|A/(2\pi \hslash )$. Adjacent states within a formerly degenerate level have the electric-field splitting of a Landau level in a rectangleThus this is the ground-to-first-excited gap when the lowest Landau band contains at least two allowed guiding centres and $\Delta E_{\mathrm{electric}}<\hslash {\omega }_c$. Without that implicit weak-field or large-sample condition, the exact full spectral gap is

Choose the Landau gaugeIt preserves translation invariance in the $y$-direction, so $p_y=\hslash k$ is conserved. With $p_z=0$, magnetic minimal coupling giveswhereThis is a quantum harmonic oscillator, so the Landau levels areindependent of $k$.

The periodic boundary condition in $y$ quantizes $k=2\pi r/L_y$. Requiring the guiding centre to lie in a strip of width $L_x$ gives a $k$-interval of width $|qB|L_x/\hslash$. The number of allowed integers is therefore approximatelythe degeneracy of a Landau level, up to boundary corrections.

No. Consider two uncoupled oscillators with nonzero amplitudes:The orbit is bounded. If it had a period $T>0$, then ${\omega }_{1}T$ and ${\omega }_{2}T$ would both be integer multiples of $2\pi$, forcing ${\omega }_1/{\omega }_2\in \mathbb{Q}$. Choosing an irrational frequency ratio gives an incommensurate harmonic-oscillator orbit, which is quasiperiodic on an invariant torus and not periodic.

Write the Hamiltonian as the sum of separated one-dimensional energiesEach $(q_i,p_i)$ system is a simple harmonic oscillator and is integrable away from $(q_i,p_i)=(0,0)$ by part (b). The separated energies satisfyand their differentials are independent whenever every oscillator pair is nonzero. The Poisson-commuting separated energies therefore make the $n$-dimensional oscillator integrable. The exceptional coordinate strata have measure zero, so this proves Liouville integrability in its usual almost-everywhere sense; the origin is the only point where $dH$ itself vanishes.

Let $F_1,\dots ,F_n$ be commuting independent integrals for $(M,H)$ and let $G_1,\dots ,G_m$ be such integrals for $(\tilde{M},\tilde{H})$. Regard each as a function on the product phase space. ThenThe $F_i$ commute among themselves, the $G_a$ commute among themselves, and every cross bracket $\{F_i,G_a\}$ vanishes because the functions depend on disjoint canonical coordinates. Their differentials form two independent coordinate blocks, so all $n+m$ are independent on the product of the two regular sets. Thus the combined system is integrable, an instance of the direct product of integrable Hamiltonian systems.

For one degree of freedom, take $F_1=H$. Antisymmetry of the Poisson bracket gives $\{H,H\}=0$, so energy is conserved. The stated condition is exactly $dH\ne 0$, which supplies the required functional independence. Hence the system is integrable at every such regular point.