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Choose the Fourier series conventionFor a real potential energy, $V_{-1}=V_1^{\ast }$. The symbol $V(G_1)$ means the Fourier coefficient $V_1$, with units of energy, rather than the pointwise value of $V(x)$ at a position $G_1$.

Retain the two plane waves $|1⟩=L^{-1/2}{e}^{ikx}$ and $|2⟩=L^{-1/2}{e}^{i(k-G_1)x}$ in the nearly-free electron model. Their kinetic energies are ${ϵ}_1={\hslash }^2{k}^2/(2m)$ and ${ϵ}_2={\hslash }^2(k-G_1{)}^2/(2m)$. The matrix representation of a linear map for the Quantum Hamiltonian projected onto this two-dimensional basis isThus $\det (H-EI)=0$ givesSolving this quadratic equation yieldsChoosing the zero of potential energy so that its spatial average is $V_0=0$ reproduces the requested expression. If that average has not been removed, both energy bands are shifted by $V_0$; no band gap, group velocity or effective mass is changed by this constant shift.

The two kinetic energies coincide at $k=G_1/2$. Their coupling gives an avoided crossing with band gap $2|V_1|$. The result is exact within the projected two-wave Quantum Hamiltonian. It is an approximation to the full crystal problem, appropriate near this crossing for a weak periodic potential whose other Fourier coefficients do not cause strong mixing with remote plane waves. Small $|V_1|$ alone would not justify ignoring arbitrarily large other Fourier coefficients.

Orthogonality means zero inner product, not merely that two objects look different. For complex vectors, $⟨u,v⟩={\sum }_i{u}_i^{\ast }{v}_i$; for functions a common choice is $⟨u,v⟩=\int w(x)u(x{)}^{\ast }v(x)dx$ with positive weight $w$. An orthonormal set has $⟨u_m,u_n⟩={\delta }_{mn}$.

For an orthogonal set, the component of $f$ along $u_n$ has coefficientA complete orthonormal set gives $f={\sum }_n{c}_n{u}_n$ in the appropriate norm and Parseval's identity $\Vert f{\Vert }^2={\sum }_n|c_n{|}^2$. Without completeness there can be a nonzero residual orthogonal to the entire chosen set. Orthogonal nonzero vectors are linearly independent, but independence alone does not imply orthogonality.

Fourier coefficients are projections becauseLikewise, for a self-adjoint operator, $A{u}_n={\lambda }_n{u}_n$ gives $({\lambda }_n-{\lambda }_m)⟨u_m,u_n⟩=0$, so distinct eigenvalues have orthogonal eigenvectors. Degeneracy does not itself force a chosen pair to be orthogonal, but an orthonormal basis can be selected within each eigenspace. The differential version uses the correct Sturm-Liouville weight and boundary conditions.

For mechanical normal modes with unequal masses the relevant inner product is $⟨u,v{⟩}_{\mathsf{M}}=u^{†}\mathsf{M}v$. This makes the modes in D14 orthogonal even when an ordinary unweighted dot product does not. Expanding in such coordinates diagonalizes the kinetic and potential quadratic forms, explaining why modal motions evolve independently and their energies add.

Use two plane waves, $|\mathbf{k}⟩$ and $|\mathbf{k}-\mathbf{G}⟩$, as a basis near their Bragg degeneracy. The potential has Fourier coefficients $V_{\pm \mathbf{G}}$; its zero Fourier coefficient vanishes. All other plane waves are off resonance and are neglected in the leading nearly-free electron model. With ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m_e)$, the Schrödinger equation reduces to the Hermitian matrix problemSetting the characteristic polynomial equal to zero givesSolving this quadratic equation givesFor the real cosine potential its coefficient $V_G$ is real; the absolute-value form also allows an arbitrary choice of origin and corresponding Fourier phase.

At the Brillouin-zone boundary, $\mathbf{k}\cdot \mathbf{G}=G^2/2$, the free energies are equal. The two standing waves have different probability densities relative to the lattice potential and are split byRepeating this avoided crossing at the reciprocal lattice boundaries joins smooth branches into energy bands. In one dimension a nonzero Fourier coefficient opens a gap at the corresponding crossing. In higher dimensions a direct gap at one wavevector need not give a common forbidden energy interval at every wavevector: different bands may still overlap in energy. The present two-dimensional example remains metallic even though the crossings are locally gapped.

For the rectangular lattice, choose real-space primitive vectors $a\hat{\mathbf{x}}$ and $b\hat{\mathbf{y}}$. The reciprocal lattice vectors areDefine the half-widths of the first Brillouin zone by $K=\pi /a$ and $L=\pi /b$. Their values and the zone areA primitive reciprocal cell can be drawn with a corner at the origin; its centred translate is the first Brillouin zone here. Its area is $4{\pi }^2/(ab)=37.9600\mathrm{n}{\mathrm{m}}^{-2}$.

Including both spin states, counting the occupied free-electron states per unit area givesConsequentlyThe occupied disk has precisely the area of the first Brillouin zone, but their shapes differ: $K<k_F<L$. It protrudes through both vertical zone boundaries and does not reach the horizontal ones. Thus two electrons per cell do not imply a filled first band in the absence of a separating band gap.

Fold the protruding caps by $\pm {\mathbf{G}}_x$. The original circle and the translated circle intersect at $k_x=K$, $k_y=\pm \sqrt{k_F^2-K^2}$. The potential mixes these states and replaces each crossing by an avoided crossing. The lower branch reconnects into two open sheets running across successive zones in the $k_x$ direction; the upper branch forms a closed lens-shaped pocket about a vertical zone boundary. In a reduced-zone sketch its two halves sit at opposite edges, but periodic identification makes them one closed sheet. This is rectangular-lattice Fermi-surface reconstruction.

The largest pocket in this weak-potential construction is the lens-shaped intersection of two disks enclosed by the intersecting free circles. Each has radius $k_F$ and their centres are separated by $2K$. If $\theta =\arccos (K/k_F)$, two sectors minus their two triangles give the lens bound on a reconstructed Fermi pocket:Opening a gap raises the upper branch at fixed chemical potential and shrinks its pocket. The reported quantum oscillation area satisfiesIt is therefore consistent with this closed sheet. Onsager quantization gives $F=\hslash A_k/(2\pi e)\simeq 601\mathrm{T}$. An orbit on this closed contour can return to itself under the Lorentz force and occupy quantized Landau levels. The open sheets do not enclose a finite orbit area for a perpendicular field in this two-dimensional model. The observation is consistent with the reconstructed pocket, without uniquely determining a completely arbitrary band structure.

For a quantitative rough coupling estimate, use the supplied fixed-chemical-potential approximation and retain the two waves coupled by ${\mathbf{G}}_x$. WriteTheir upper eigenvalue isSetting it equal to $C{k}_F^2$ gives the pocket boundaryIts intersection with $k_y=0$ occurs at $q=\pm q_m$, whereThe two-wave estimate of a Bragg-gap coupling from a pocket area therefore usesAt $v=0$ this reproduces the lens bound on a reconstructed Fermi pocket; the pocket disappears at $v=C(k_F^2-K^2)=0.23786\mathrm{eV}$. Numerical integration gives $A(0.075\mathrm{eV})=6.22\mathrm{n}{\mathrm{m}}^{-2}$ and $A(0.100\mathrm{eV})=5.54\mathrm{n}{\mathrm{m}}^{-2}$. Matching the measured area gives $v\simeq 0.0931\mathrm{eV}$, so the justified rough result isThis estimate neglects remote plane-wave admixtures and other Fourier coefficients. It also holds the chemical potential fixed as requested; an exact fixed-density calculation would determine it self-consistently after reconstruction. The numerical root is a way to implement the approximate model, not a measurement of $V_G$ to four significant figures.

Why degenerate perturbation theory is needed. If several unperturbed states have equal or nearly equal energy, a perturbation can mix them at leading order. In that subspace, diagonalize the perturbation matrix rather than using nondegenerate formulas with small energy denominators. The resulting eigenvectors identify the combinations selected by the coupling, and the eigenvalues give the leading splitting. Other remote states can then be included perturbatively.

For the linear combination of atomic orbitals description of molecular hydrogen, equivalent atomic $1s$ states on the two nuclei supply the nearly degenerate starting functions. In an orthonormal two-state model, equal diagonal energy $E_a$ and real hopping $t$ give $E_a\pm t$ and symmetric/antisymmetric combinations. For actual overlapping atomic functions with overlap $S$ and Hamiltonian coupling $h$, one instead solves a generalized eigenvalue problem:The bonding state has enhanced density between the nuclei and the antibonding state a separating node. Two opposite-spin electrons occupying the bonding molecular orbital describe the simplest closed-shell hydrogen-molecule bond. Electron-electron interaction and nuclear repulsion are needed to calculate a realistic total binding energy; the two-state splitting alone is not the full molecular energy.

In the nearly-free electron model, the analogous degenerate states are plane waves $|\mathbf{k}⟩$ and $|\mathbf{k}-\mathbf{G}⟩$ at a Bragg scattering. A Fourier coefficient $U_{\mathbf{G}}$ of the periodic potential couples them. At exact degeneracy $E_B$, diagonalizing their two-by-two matrix givesThe associated standing waves place electronic density differently relative to the ionic potential. These are the same gap-opening combinations illustrated in Q1(b). The labels of the two starting basis states survive as mixed orbital or plane-wave character, rather than independent crossing eigenstates.

Origin of broad and narrow bands. Spatially extended s/p-derived orbitals have substantial overlap and therefore large hopping amplitudes and broad bands. More localized d orbitals overlap less strongly and produce narrower bands. A multiorbital tight-binding calculation assigns the on-site orbital energies and hopping matrices, forms Bloch sums, and diagonalizes the resulting matrix at each wavevector. Different hopping ranges and orbital symmetry determine the bandwidths and allowed interband coupling. The two-band description retains one broad combination and one nearly flat d-derived combination; a real transition metal contains several d orbitals and more complicated dispersions.

Hybridized eigenvalues and eigenstates. Let $|{\psi }_{\mathbf{k}}⟩=a|{\varphi }_{\mathbf{k}}⟩+b|{\chi }_{\mathbf{k}}⟩$, in an orthonormal basis. Hermitian operator fixes the reverse coupling to be $V^{\ast }$. Projection onto the two components givesSolving this quadratic yields the band hybridization dispersionFor nonzero $V$, one eigenvector representation is $(a,b)\propto (V,E-E_1)$, followed by normalization. Thus the wavefunctions exchange broad-band and d-band character through the avoided crossing; the levels are equal mixtures at the uncoupled crossing.

At $V=0$, the parabola crosses the horizontal level at $k=\pm k_0$, with $k_0=\sqrt{2m{E}_d}/\hslash$. For small nonzero $|V|$, the curves avoid crossing with minimum direct separation $2|V|$. The lower branch follows the parabola below the crossing and flattens toward $E_d$ above it; the upper does the opposite. The lower branch satisfies $E_{-}<E_d$ everywhere and tends to $E_d$ from below as $k\to \infty$, while the upper branch always lies above $E_d$. This is the perfectly flat-band continuum approximation, with zone-boundary intersections excluded as requested.

Fixed particle number and the larger Fermi sphere. Denote the uncoupled broad-band Fermi radius by $k_0$ to distinguish it from the new one $K$. With two spin states, the broad-band number density at $V=0$ is $k_0^3/(3{\pi }^2)$. The stipulated equal population in the flat sector makes the total density twice that value. At zero temperature and any nonzero hybridization, all occupied states fit into the monotonic lower branch below $E_d$, while the upper branch is empty. ThereforeThis is the stated factor increase. It uses fixed total particle number, equal spin counting, an exactly flat unhybridized band and enough available states below any Brillouin-zone restriction. In generic pairs of dispersive bands both hybridized pockets can remain occupied; then Fermi-volume conservation in two hybridized electron bands fixes a sum of volumes rather than this one radius. The exactly flat band is degenerate at $V=0$, so its initial occupation must be specified, as it is here.

Fermi-velocity suppression. Set $\Delta (k)=E_1(k)-E_d$. The exact derivative of the lower branch isAt $k=K$, $E_1(K)=2^{2/3}{E}_d$ and $\Delta (K)=(2^{2/3}-1)E_d>0$. For $|V|\ll \Delta (K)$, the square bracket divided by two is $|V{|}^2/\Delta (K{)}^2+O(|V{|}^4/\Delta (K{)}^4)$. Since group velocity is ${\hslash }^{-1}dE/dk$ and the broad-band reference velocity is $v_F^{(1)}=\hslash k_0/m$,so the flat-band hybridization Fermi-velocity suppression isThe reference velocity is that of the original broad-band Fermi sphere, not the broad-band slope at the larger $K$; omitting the radius change loses the factor $2^{1/3}$.

The low-temperature qualification becomes more stringent as $V\to 0$: the new chemical potential approaches $E_d$ with $E_d-\mu \simeq |V{|}^2/\Delta (K)$, and the relevant thermal energy must be small compared with this hybridization scale. At fixed nonzero temperature the singular flat-band redistribution is not automatically resolved for arbitrarily small coupling. A small but finite narrow-band dispersion can also change the occupation topology.

Density of states and magnetism. For an isotropic three-dimensional branch with spin degeneracy two,A small Fermi velocity therefore produces a large density of states. The localized, partially filled d-derived bands in iron, cobalt and nickel place many states close to the Fermi level; broad s/p-derived conduction bands in copper, silver and gold coexist with filled d bands lying below it. This explains the substantially larger Fermi-level density of states of the former group within the band picture. An exact factor of ten requires the actual band parameters and is not fixed by the two-band toy model.

A high density of states lowers the kinetic-energy cost of spin polarization. A repulsive exchange interaction can then satisfy the Stoner criterion and stabilize ferromagnetism. With a single-spin density of states $g_s$ and an on-site interaction normalization, the criterion is $U{g}_s(E_F)>1$; with spin-summed $g=2g_s$, it is $Ug(E_F)>2$. The large d-state contribution favors this instability in Fe, Co and Ni. The noble metals do not satisfy the corresponding instability condition in the same ordinary state; this does not imply that they have no weak magnetic response whatsoever.

Both models respect Bloch's theorem and describe electrons in a periodic lattice, but they start from different useful limits.

In the nearly-free electron model, start with extended plane waves and free-electron energies ${\epsilon }_{\mathbf{k}}^0={\hslash }^2{k}^2/(2m)$. A weak periodic potential has Fourier coefficients $V_{\mathbf{G}}$ at reciprocal lattice vectors and couples $\mathbf{k}$ to $\mathbf{k}-\mathbf{G}$. Away from degeneracies its corrections are small. Near a Bragg scattering, retain the two resonant plane waves and diagonalizeThe resulting eigenvalues areAt the crossing the gap is $\boxed{2|V_{\mathbf{G}}|}$. The two standing-wave combinations place electronic density differently relative to the periodic potential and therefore acquire different energies. Folding the many free parabolas into the first Brillouin zone produces the repeated band labels; the nonzero potential opens avoided crossings between coupled states. A Fourier component forbidden by symmetry need not open its corresponding first-order gap.

In the tight-binding model, instead start with localized orbitals on well-separated atoms or sites, with discrete on-site energies. Their overlap permits tunneling represented by hopping integrals. Bloch sums of these orbitals diagonalize the lattice translation symmetry. For one orthonormal orbital per site in a one-dimensional nearest-neighbor chain,The many initially degenerate copies of one atomic orbital broaden into a band. Other atomic orbitals or multiple sites per primitive cell produce additional bands; their hybridization creates bonding and antibonding combinations and may open gaps. For two orbitals per cell, one illustrative Hamiltonian iswith eigenvalues $(E_A+E_B)/2\pm \sqrt{(E_A-E_B{)}^2/4+|t_1+t_2{e}^{-ika}{|}^2}$. This shows explicitly how two bands arise from the two cell degrees of freedom, even when the isolated sites have equal on-site energies.

For each orbital degree of freedom per primitive cell there is one band per spin; band filling and the presence of a gap at the Fermi level determine metallic or insulating behavior. The nearly-free limit emphasizes weak Bragg scattering of extended states, while tight binding emphasizes hopping between localized states. They are complementary approximations to electronic band structure, rather than two incompatible explanations of why periodic solids have bands.

For a lattice-periodic one-electron Hamiltonian, Bloch's theorem givesHere $\mathbf{R}$ is any lattice translation and $n$ labels an energy band. Under that translation the wavefunction gains phase $e^{i\mathbf{k}\cdot \mathbf{R}}$. The crystal momentum $\hslash \mathbf{k}$ labels this translation eigenvalue and is defined modulo $\hslash \mathbf{G}$ for a reciprocal lattice vector $\mathbf{G}$.

The mechanical momentum is different. In zero magnetic field its operator is $-i\hslash \nabla$, which acts asA general Bloch state is therefore not a mechanical-momentum eigenstate. Expanding its periodic factor gives plane-wave components with momenta $\hslash (\mathbf{k}+\mathbf{G})$, with probabilities determined by their Fourier coefficients. For a local Hamiltonian $p^2/(2m_e)+V(\mathbf{r})$, its mean momentum satisfiesnot generally $\hslash \mathbf{k}$. The lattice exchanges actual momentum with the electron; crystal-momentum conservation in scattering is instead modulo reciprocal-lattice vectors, as in Umklapp scattering.

The Moiré pattern sets a superlattice periodFor the stipulated scalar one-dimensional model, right- and left-moving states have $E(k)=\hslash v|k|$. The $n$th Bragg crossing satisfies $E(k)=E(k-nG)$ and hence $k=nG/2$. The Bragg gap ladder for a linear dispersion therefore has centresThis is the spacing of successive possible Bragg gaps. The corresponding atomic-lattice scale is $hv/(2a)\simeq 8.27\mathrm{eV}$, fifty times larger.

Shifting a linear branch by one full reciprocal period $G$ instead gives $\hslash vG\simeq 0.331\mathrm{eV}$, twice the consecutive Bragg-threshold spacing; the first zone boundary is at $G/2$. These scales should not be confused with the gap width, which depends on the weak potential's Fourier coefficients. The geometry alone does not guarantee equally wide or even observable gaps at every threshold, nor an exactly periodic full spectrum. The estimate neglects graphene's additional pseudospin selection rules and interlayer correlations, as required by the scalar-chain approximation.

Interpret the source's missing variable in the limit arrow as $t\to \infty$. We prove convergence in distribution by showing that the angle of the Brownian motion becomes uniform on the circle group.

Let $Z_t=e^{iB(t)}$. For every integer $m$, its angular Fourier coefficients satisfyThese are precisely the Fourier coefficients of the uniform distribution on the unit circle. Thus expectations of every trigonometric polynomial converge to its uniform angular average. By the Stone-Weierstrass theorem, trigonometric polynomials approximate continuous functions on the unit circle uniformly. Given an approximation error $\epsilon$, the discrepancy of expectations is at most the polynomial discrepancy plus $2\epsilon$. Taking $t\to \infty$ and then $\epsilon \to 0$ proves weak convergence of probability measures to uniform angular measure.

The function $z\mapsto (\mathrm{Im}z{)}^2$ is continuous on the unit circle, so the continuous mapping theorem and part (a)(ii) implyEquivalently, the limiting arcsine distribution has cumulative distribution function $2\arcsin \sqrt{x}/\pi$ for $0\le x\le 1$, and probability density function $1/(\pi \sqrt{x(1-x)})$ for $0<x<1$. This proves the limiting law rather than inferring it merely from the limiting mean.

A summability kernel on the circle is a sequence of continuous periodic functions withThe frequently used stronger condition of uniform convergence to zero off each neighborhood of zero implies the last condition. For $f$ in a homogeneous Banach space,Given $\epsilon >0$, translation continuity bounds the difference inside a sufficiently small neighborhood by $\epsilon$; its integral is at most $C\epsilon$. Outside, use $2\Vert f{\Vert }_B$ and the vanishing tail condition. Hence $f\ast k_n\to f$ in $B$.

With the indexing needed in part (c), the Fejér kernel isIn particular $K_0=1$. The squared-modulus expression proves $K_n\ge 0$, and its constant Fourier coefficient proves $\int K_{n}dm=\Vert K_n{\Vert }_1=1$. The geometric series yieldsThus these are summability kernels. For $T_f:g\mapsto f\ast g$, part (a) gives $\Vert T_f\Vert \le \Vert f{\Vert }_B$. Since $\Vert K_n{\Vert }_1=1$ and $T_f{K}_n\to f$ in $B$, the reverse bound follows:

For $n\ge 1$ put $q_n(t)=2n{K}_{n-1}(t)\sin (nt)$. Using $\sin (nt)=(e^{int}-e^{-int})/(2i)$ and the Fourier coefficients of $K_{n-1}$ givesIndeed, when $1\le k\le n$ only the $e^{int}$ shift contributes, with coefficient $(-in)(1-|k-n|/n)=-ik$. For $-n\le k\le -1$ only the opposite shift contributes and gives $in(1-|k+n|/n)=-ik$; at $k=0$ both coefficients vanish.

If $P={\sum }_{|k|\le n}{a}_k{e}^{ikt}$, convolution multiplies its Fourier coefficients by those of $q_n$. Thus $-P\ast q_n=\sum ik{a}_k{e}^{ikt}=P^{\prime }$, proving the exact identity. Since $K_{n-1}\ge 0$ and has integral one,For $n=0$ the polynomial is constant and the bound is immediate. This elementary Fejér-kernel argument proves the requested estimate; the sharp Bernstein inequality for trigonometric polynomials has constant $n$.

Put $z=e^{i\theta }$, so $d\theta =dz/(iz)$, $\cos 2\theta =(z^2+z^{-2})/2$ and ${\cos }^{2}3\theta =(z^6+2+z^{-6})/4$. The contour integral isThe enclosed poles are $0$ and $\pm 1/\sqrt{2}$. For the rational function displayed without the prefactor $1/(4i)$, the residue at zero is $-21/8$ and the two other residues sum to $27/8$. Therefore the residue theorem givesA useful independent check is the Fourier series $(5-4\cos \varphi {)}^{-1}=\frac{1}{3}[1+2{\sum }_{k\ge 1}{2}^{-k}\cos k\varphi ]$; only its constant and third Fourier coefficients survive.

Put $z=e^{i\theta }$, so $d\theta =dz/(iz)$, $\cos 2\theta =(z^2+z^{-2})/2$ and ${\cos }^{2}3\theta =(z^6+2+z^{-6})/4$. The contour integral isThe enclosed poles are $0$ and $\pm 1/\sqrt{2}$. For the rational function displayed without the prefactor $1/(4i)$, the residue at zero is $-21/8$ and the two other residues sum to $27/8$. Therefore the residue theorem givesA useful independent check is the Fourier series $(5-4\cos \varphi {)}^{-1}=\frac{1}{3}[1+2{\sum }_{k\ge 1}{2}^{-k}\cos k\varphi ]$; only its constant and third Fourier coefficients survive.

For a nonnegative integer $m$, expand the plane wave in powers of $\rho$:The first contribution to the positive angular harmonic $e^{im\theta }$ occurs at $n=m$. Its coefficient is $(i\rho {)}^m/(2^{m}m!)$. Comparing this with $a_m{J}_m(\rho )\sim a_m{\rho }^m/(2^{m}m!)$ gives $a_m=i^m$ for $m\ge 0$.

The plane wave is even in $\theta$, so its $m$ and $-m$ Fourier coefficients agree. Therefore $a_{-m}{J}_{-m}=a_m{J}_m$, implying $a_{-m}=(-1{)}^m{i}^m=i^{-m}$. HenceThis is the cosine form of the Jacobi-Anger expansion. The factorial asymptotic stated in the source is used for nonnegative order; applying it literally to negative integers would introduce an undefined factorial. The Integer-order Bessel reflection identity is the required extension, not a separate set of divergent radial solutions.

Subtract the particular plane Poiseuille flow profile from the full rectangular duct flow. The remainder is a harmonic function:The data are even in $x$ and $y$. The boundary value problem is unique: if two solutions differed by $v$, Green's first identity and the zero Dirichlet boundary condition would giveThus $v$ is constant, and its zero boundary values force $v=0$. Reflection of a solution in either coordinate axis therefore gives the same solution. We may seek even separated eigenfunctions.

Use separation of variables withEach summand solves the Laplace equation, vanishes at $y=\pm b$, and has the required evenness. The hyperbolic cosine ratio equals one at either side wall. The functions $\cos (k_{n}y)$ have orthogonality relationThey are the even Dirichlet eigenfunctions on $[-b,b]$, so their Fourier series represents the even boundary data. Projecting that data onto the $n$th eigenfunction yields the Fourier coefficientFor completeness, two integrations by parts give the quadratic arch expansion in half-integer cosine modes: the displayed integral is $4(-1{)}^n/k_n^3$, agreeing with the identity supplied in the paper.

Adding the particular solution givesThe coefficients are absolutely summable and the hyperbolic cosine ratio is at most one throughout the closed rectangle, giving uniform convergence and the prescribed boundary values. On compact subsets of the interior, the hyperbolic cosine ratio supplies exponential decay in $n$, justifying differentiation of the series and the Laplace equation. The side-wall correction slows the plane Poiseuille flow enough to satisfy all four no-slip boundary conditions.

Use the period-two Fourier series conventionAway from its jumps this square wave is an odd function, so $a_0=a_n=0$. Its Fourier sine coefficients areOnly odd Fourier harmonics remain. ThereforeThe Fourier series equals $f(x)$ away from the jumps; at a jump it gives the midpoint of the one-sided limits by Dirichlet's theorem on Fourier series. The printed definition leaves the values at the integers unspecified. Choosing finite values there consistent with period two changes neither the Fourier coefficients nor these limiting values.

The given square wave agrees almost everywhere with an odd function on $(-\pi ,\pi )$; its assigned value at the single point $0$ does not affect any Fourier coefficient. Thus $a_0=a_n=0$, whileThe Fourier series is thereforeThe Dirichlet convergence theorem for Fourier series applies because the periodic extension is piecewise smooth. It gives $f(x)$ at each continuity point and the midpoint $0$ at the jumps, including $x=0$; it does not reproduce the assigned value $f(0)=-1$ there.

At the continuity point $x=\pi /2$, $\sin ((2r+1)\pi /2)=(-1{)}^r$ and $f(\pi /2)=1$. ThusNo evaluation at a discontinuity or rearrangement of the conditionally convergent series is involved.

The function is even, so every sine coefficient is zero. Its mean coefficient isFor $n\ge 1$, compute $a_n=(2/\pi ){\int }_0^{\pi }\sin x\cos (nx)\mathrm{d}x$. The product-to-sum identity makes this integral zero for odd $n$ and gives $-4/(\pi (n^2-1))$ for even $n$. Thus the missing Fourier coefficients are $a_n=0$ for odd $n$ and $b_n=0$ for every $n$, with $a_0=4/\pi$. Equivalently,The Fourier series converges to this continuous periodic function; there is no jump value to average.

Seek a periodic particular solution with Fourier coefficients $A_n,B_n$. Comparing constant terms gives $A_0=a_0/2$. Comparing the cosine and sine terms for $n\ge 1$ givesso $A_n=2a_n/(n^2+4)$ and $B_n=n{a}_n/(n^2+4)$. Insert the coefficients from part (a) and add the homogeneous solution:The differentiated series converges absolutely and uniformly, because its largest coefficient terms are $O(m^{-2})$. This justifies the term-by-term first derivative used in substituting into the differential equation. Any two solutions differ by $C{e}^{-2x}$, proving that the displayed family is the general solution. The computation is an instance of Fourier response of a first-order linear differential equation.