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The acoustic pressure satisfies $p_{tt}=c_0^2\Delta p$. Rigid walls impose vanishing normal velocity, equivalently the Neumann boundary condition on pressure. Separation of variables gives transverse modes $\cos (m\pi x/a)\cos (n\pi y/a)$ with $m,n\ge 0$. With axial factor $e^{ikz-i\omega t}$, their dispersion relation isThe $(0,0)$ mode is nondispersive. Every other mode propagates only above its waveguide cutoff frequency ${\omega }_{mn}=\pi c_0\sqrt{m^2+n^2}/a$. Thus ${\omega }_{\min }=\pi c_0/a$.

Expand the prescribed entrance pressure in this cosine basis, with coefficients $A_{mn}$. Impose outgoing propagation and decay at infinity, with no incoming disturbance. At $\omega ={\omega }_{\min }/2$, only the constant transverse mode propagates, with $k_{00}=\pi /(2a)$. All other modes have $k=i{\chi }_{mn}$, where ${\chi }_{mn}=\pi \sqrt{m^2+n^2-1/4}/a$. The exact modal field is the outgoing constant mode plus ${\sum }_{m+n>0}{A}_{mn}\cos (m\pi x/a)\cos (n\pi y/a)e^{-{\chi }_{mn}z-i\omega t}$.

Since the slowest evanescent decay rate is $\sqrt{3}\pi /(2a)$, the far field isIf that area average is zero, there is no propagating far-field component, and the field decays exponentially instead. For sufficiently regular entrance data the omitted terms are $O(e^{-\sqrt{3}\pi z/(2a)})$.

The rate-of-strain tensor is $e_{ij}=\frac{1}{2}({\partial }_j{u}_i+{\partial }_i{u}_j)$. For an incompressible Newtonian fluid, the stress tensor is ${\sigma }_{ij}=-p{\delta }_{ij}+2\mu e_{ij}$. Dotting the momentum equation with velocity and using the product rule gives the local mechanical-energy balancewithout body forces. Write ${\partial }_j{u}_i=e_{ij}+{\omega }_{ij}$, with ${\omega }_{ij}$ antisymmetric. The pressure contribution is $-p{\partial }_i{u}_i=0$, and $e_{ij}{\omega }_{ij}=0$ by symmetry. Thus the remaining loss is $2\mu e_{ij}{e}_{ij}\ge 0$, proving the local viscous dissipation rate is $2\mu e_{ij}{e}_{ij}$ per unit volume.

For the oscillating boundary, take $u=(U(y,t),0,0)$ and constant pressure. The nonlinear advection vanishes, leaving $U_t=\nu U_{yy}$, $\nu =\mu /\rho$. Seek the real part of $V{e}^{i\omega t-\gamma y}$; decay at infinity selects $\gamma =(1+i)/\delta$, $\delta =\sqrt{2\nu /\omega }$. After transients,Only $e_{xy}=e_{yx}=U_y/2$ is nonzero, so the dissipation is $\mu U_y^2$. Its time average is $\mu V^2{\delta }^{-2}{e}^{-2y/\delta }$: the two sine and cosine contributions each average to one half. Integrating over $y>0$ gives

Use metric signature $(+---)$ and write $u^a={\dot{x}}^a$, $U=\sqrt{{\eta }_{ab}{u}^a{u}^b}$. For an orientation-preserving affine change $s^{\prime }=\alpha s+\beta$, with $\alpha >0$, $u^{\prime a}=u^a/\alpha$ and $d{s}^{\prime }=\alpha ds$. Homogeneity of both terms of the Lagrangian therefore leaves $Lds$ unchanged. Proper-time parametrization is used again when deriving the final equation.

Under $A_a\mapsto A_a+{\partial }_a\chi$, the action changes by $-q\int d\chi =-q[\chi (x_f)-\chi (x_i)]$. The fixed-endpoint variational principle is gauge invariant; the action itself can change by a boundary term. This is the precise version of the printed invariance claim. Negative $\alpha$ reverses orientation and requires a corresponding reorientation of the path, rather than the simple positive-parameter calculation above.

The canonical momentum isThe Euler-Lagrange equation gives $-m\frac{d}{ds}(u_a/U)-q{\partial }_b{A}_a{u}^b=-q{\partial }_a{A}_b{u}^b$. Hence, with $F_{ab}={\partial }_a{A}_b-{\partial }_b{A}_a$ and proper time $U=1$,Antisymmetry verifies the consistency condition directly: $\frac{d}{ds}(u_a{u}^a)=2u_a{\dot{u}}^a=(2q/m)u^a{F}_{ab}{u}^b=0$.

For the uncharged Lagrangian in the same sign convention, $p_a=-m{u}_a/U$, so $P_a=p_a-q{A}_a$. Equivalently $P_a+q{A}_a$ is the gauge-invariant kinetic covector in this convention. This illustrates electromagnetic minimal coupling: replace the free canonical momentum by the gauge-covariant combination when introducing the field. The physical contravariant kinetic four-momentum is conventionally $m{u}^a$; its sign must not be confused with the covector obtained by differentiating the particular negative-length Lagrangian used here.

The chemical potential is $\mu =(\partial E/\partial N{)}_{S,V}$, equivalently $(\partial F/\partial N{)}_{T,V}$: it is the free-energy cost of adding a particle at fixed temperature and volume. With units $k_B=1$, a large reservoir has entropy $S_R(E_{\mathrm{tot}}-E,N_{\mathrm{tot}}-N)=S_R^0-E/T+\mu N/T$ to first order. Since the number of compatible reservoir states is proportional to its exponential, the grand canonical ensemble assignsFor independent fermions, a one-particle state of energy $\epsilon$ has occupation zero or one. Its partition factor is $1+e^{-(\epsilon -\mu )/T}$, so the mean occupation is the Fermi-Dirac distribution $[e^{(\epsilon -\mu )/T}+1{]}^{-1}$. Multiplication by the given density of states yields $dN=C{\epsilon }^{1/2}d\epsilon /[e^{(\epsilon -\mu )/T}+1]$.

At zero temperature the occupation is a step function, filled below $\mu$ and empty above it, so $\mu ={\epsilon }_F$ and $N=(2C/3){\epsilon }_F^{3/2}$. At fixed particle number, the low-temperature shift of $\mu$ is $O(T^2/{\epsilon }_F)$ and does not affect the leading estimate. In the narrow interval of width $T$ above ${\epsilon }_F$, replace the density of states by $C\sqrt{{\epsilon }_F}$ and put $u=(\epsilon -{\epsilon }_F)/T$. ThenEquivalently, the fraction above the Fermi energy is asymptotic to $(3\log 2/2)T/{\epsilon }_F$.

There is a genuine periodicity mismatch in the source. Write $x_s=sb+u_s$. The intended periodic displacements in a harmonic chain satisfy $u_N=u_0$, equivalently $x_N=x_0+Nb$ for unwrapped positions. The printed $x_N=x_0$ instead makes the potential $\frac{1}{2}m{\lambda }^2{\sum }_s(x_{s+1}-x_s{)}^2+\frac{1}{2}Nm{\lambda }^2{b}^2$, with equilibrium at equal positions, not $x_s=sb$. The following lattice derivation uses periodic displacements, as required by the requested position operator.

Put $k_r=2\pi r/(Nb)$, $r=0,\dots ,N-1$, and Fourier transform the displacements and momenta with normalization $N^{-1/2}$. The spring quadratic form is diagonal, with factor $|e^{i{k}_{r}b}-1{|}^2=4{\sin }^2(k_{r}b/2)$. Thus each nonzero mode is a harmonic oscillator withCreation and annihilation operators obey $[a_r,a_{r^{\prime }}^{†}]={\delta }_{r{r}^{\prime }}$. Inverting the oscillator-coordinate expansion givesThe conjugate momentum is $p_s=-i\sqrt{m\hslash /(2N)}{\sum }_{r\ne 0}\sqrt{{\omega }_r}(a_r{e}^{i{k}_{r}sb}-a_r^{†}{e}^{-i{k}_{r}sb})$. Fourier orthogonality diagonalizes the Hamiltonian as ${\sum }_{r\ne 0}\hslash {\omega }_r(a_r^{†}{a}_r+1/2)$. The normalized oscillator vacuum is annihilated by every $a_r$; applying creation operators gives number states, each added phonon contributing $\hslash {\omega }_r$. The omitted zero mode is a free centre-of-mass coordinate, which must be fixed or treated separately rather than assigned an oscillator vacuum. Accordingly the reduced displacement commutator is $[u_s,p_t]=i\hslash ({\delta }_{st}-1/N)$.

For $iq{u}_s$, separate creation and annihilation parts. Their commutator is a scalar, so the supplied Baker--Campbell--Hausdorff formula and vacuum expectations giveSumming the geometric lattice phases yieldsThe ratio is interpreted by continuity at its removable singularities. The interference factor has its maximum $N^2$ at $q=2\pi n/b$, where all lattice phases coincide. The source's Fourier-sum hint should read $r\equiv 0(\mathrm{mod}N)$, rather than $r=Nb$.

The claim about the full matrix element also needs qualification: its Debye–Waller factor envelope varies with $q$. The absolute global maximum is $|M(0){|}^2=N^2$; at nonzero reciprocal-lattice points the value is $N^2{e}^{-2W(2\pi n/b)}<N^2$. Moreover the derivative of the envelope there is nonzero, whereas that of the interference factor is zero, so those points are not exact local maxima of the full finite-chain intensity. The reciprocal-lattice condition maximizes the interference factor, which is the usual constructive-scattering interpretation of the requested result.

Expand a perturbed eigenvector and energy as $\psi ={\psi }_0+\lambda {\psi }_1+O({\lambda }^2)$ and $E(\lambda )=E+\lambda \delta +O({\lambda }^2)$. At first order, $(H-E){\psi }_1+(V-\delta ){\psi }_0=0$. Projecting onto the $E$ eigenspace eliminates the first term and gives $P_{E}V{P}_E{\psi }_0=\delta {\psi }_0$. Thus degenerate perturbation theory diagonalizes the restriction of $V$ to that eigenspace; in the nondegenerate case this reduces to its expectation value.

For the specified two-dimensional eigenspace, the restricted matrix is $\left(\begin{array}{cc}\alpha & \beta \\ \beta & \alpha \end{array}\right)$. Its eigenvectors are $(|1⟩\pm |2⟩)/\sqrt{2}$ and its eigenvalues $\alpha \pm \beta$. Hence the perturbed energies are $E+\lambda (\alpha \pm \beta )+O({\lambda }^2)$.

The square-box unperturbed energies are $E_{pq}={\pi }^2{\hslash }^2(p^2+q^2)/(2m{a}^2)$. Every diagonal matrix element of $xy/a^2$ is $1/4$, because each coordinate has mean $a/2$. The ground state is $(p,q)=(1,1)$, giving $E_{11}+\lambda /4$. The first excited eigenspace is spanned by $(1,2)$ and $(2,1)$. Its off-diagonal entry factors into the square ofThus that entry is $256/(81{\pi }^4)=(4/(3\pi ){)}^4$, and the three levels areThe neglected energy corrections are $O({\lambda }^{2}m{a}^2/{\hslash }^2)$ in the given weak-field regime.

Put $t=\sqrt{z}\tau$ and $L=z^{3/2}$. The steepest descent phase is $F(\tau )=-{\tau }^3/3+\tau$, with saddles at $\pm 1$. The specified contour can be deformed through the negative saddle, where $F(-1)=-2/3$; the positive saddle would produce the growing exponential and is not the saddle for this contour.

An explicit descending contour is $\tau =u+iv$, where $u=-\sqrt{1+v^2/3}$ and $v$ increases from negative to positive infinity. Its imaginary phase is $v(1-u^2+v^2/3)=0$, and its ends approach the prescribed rays $4\pi /3$ and $2\pi /3$. The real part of the phase has a strict maximum at $v=0$, with quadratic term $-v^2$. Since the integrand is entire and decays in the connecting sectors, contour deformation introduces no residue.

Near the saddle, write $\tau =-1+is+O(s^2)$. Then $F(\tau )=-2/3-s^2+O(s^3)$ and $d\tau =i(1+O(s))ds$. The Gaussian saddle contribution is thereforeThe remainder of the descending contour has smaller real phase, so does not alter this leading term.

Introduce $p_{-1}=1$, $q_{-1}=0$, $p_0=a_0$ and $q_0=1$. Composition of the Möbius transformations $z\mapsto a_j+1/z$ represents a finite continued fraction. Their matrices giveIndeed, multiplying the last factor gives the linear recurrences $p_n=a_n{p}_{n-1}+p_{n-2}$ and $q_n=a_n{q}_{n-1}+q_{n-2}$, which reproduce the continued fraction convergents of the continued fraction. Each factor has determinant $-1$. Taking determinants therefore yieldsThis identity also proves that the numerator and denominator in the first column are coprime, so no unnoticed cancellation changes the displayed matrix.

Let $H_t(x)=(4\pi t{)}^{-n/2}{e}^{-|x{|}^2/(4t)}$ be the heat kernel, and write $P_{t}g=H_t\ast g$. The homogeneous solution is $u(t,x)=P_{t}g(x)$: differentiation verifies the heat equation, and the approximate-identity property recovers the smooth initial data.

The Duhamel principle givesEach source impulse at time $s$ evolves by the homogeneous heat flow for duration $t-s$. More formally, differentiating the time integral gives $f(t,x)+{\int }_0^t\Delta P_{t-s}f(s)(x)ds$. Subtracting its Laplacian leaves $f$. The integral tends to zero as $t\downarrow 0$, since its absolute value is at most $t{\sup }_{s\le t}\Vert f(s){\Vert }_{\infty }$. Thus the initial condition is also correct. Smoothness permits differentiation locally; standard cutoff or semigroup arguments justify it for bounded smooth data without global bounds on all derivatives.

For $n=4$ and time-independent Schwartz $f$, integrate the kernel in time. For $r>0$, ${\int }_0^t(4\pi s{)}^{-2}{e}^{-r^2/(4s)}ds=e^{-r^2/(4t)}/(4{\pi }^2{r}^2)$. ConsequentlyThe forcing term converges by dominated convergence to $W(x)=(4{\pi }^2|\cdot {|}^2{)}^{-1}\ast f$. Its singularity is locally integrable in four dimensions, and the Schwartz decay controls infinity. The supplied fundamental solution identity gives $-\Delta W=f$, first in distributions and then classically by regularity.

The asserted large-time limit is not guaranteed for arbitrary bounded smooth $g$. Take $f=0$ and $g(x)=(1+|x{|}^2{)}^i$, which is smooth and has modulus one. For a standard four-dimensional Gaussian $Z$,Dominated convergence gives the asymptotic, and the gamma factor is nonzero. The phase $\log t$ rotates indefinitely, so there is no limit. This is an example of heat evolution of bounded data need not converge at large times.

Under the natural additional assumption $g\in C_0({\mathbb{R}}^4)$, $P_{t}g\to 0$: split the convolution into a large ball, whose heat-kernel mass tends to zero, and its complement, where $g$ is uniformly small. Then $w=W$ and $-\Delta w=f$. More generally, if $g(x)\to L$ at spatial infinity, the limit is $w=L+W$, satisfying the same equation. These are the intended qualified stationary-limit conclusions.

The standard positive-cost interpretation requires $a>0$ and $a_0\ge 0$. These signs are not printed. Without them the claim of a unique positive fixed point is false: $a=a_0=\lambda =0$ gives $f(z)=0$, with no positive fixed point. The following derivation gives the intended result under those cost assumptions; $b_0$ may be any finite constant.

Let $V_n(x)$ be the value with $n$ stages remaining. Starting from $V_0(x)=\frac{1}{2}{a}_0{x}^2+b_0$, the Bellman equation isIf $V_{n-1}(x)=\frac{1}{2}{a}_{n-1}{x}^2+b_{n-1}$, the minimized expression is a quadratic in $u$ with positive leading coefficient $1+\beta a_{n-1}$. Its derivative vanishes at $u=k_{n}x$, and completing the square givesInduction establishes the value form and the control $u_j=k_{T-j}{X}_j$. Only zero mean and variance ${\sigma }^2$ of the independent noise were needed.

The Riccati recurrence fixed-point equation becomesThe product of its roots is $-a/\beta <0$, so exactly one is positive:For $z\ge 0$, $f^{\prime }(z)=\beta {\lambda }^2/(1+\beta z{)}^2\ge 0$, while $f(z)-z$ is positive below $a_{\ast }$ and negative above it. Monotonicity of $f$ prevents overshooting the fixed point. Thus $a_n$ increases to $a_{\ast }$ if $a_0<a_{\ast }$, decreases to it if $a_0>a_{\ast }$, and is constant if $a_0=a_{\ast }$. Continuity identifies the limit uniquely.

Iteration of the second recurrence gives $b_n={\beta }^n{b}_0+\frac{1}{2}\beta {\sigma }^2{\sum }_{j=0}^{n-1}{\beta }^{n-1-j}{a}_j$. Subtract the same expression with every $a_j$ replaced by $a_{\ast }$. The early finite part vanishes geometrically, and the later part is bounded by ${\sup }_{j\ge J}|a_j-a_{\ast }|/(1-\beta )$ times the fixed factor. Taking $n\to \infty$, then $J\to \infty$, proves