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Let $A_s$ denote the surface area of an n-sphere for the unit sphere $S^s$. In this notation $s$ is the intrinsic dimension of the sphere. Cartesian factorization of the Gaussian integral gives $I_N={\pi }^{N/2}$. In hyperspherical coordinates,The substitution $t={\rho }^2$ supplies the defining Gamma function integral. Scaling the sphere radius multiplies its surface volume by $r^{N-1}$, henceThis is the volume of the sphere itself, not of its enclosed ball.

Take $D$ to denote spacetime dimension and $d=D-1$ spatial dimensions. In uncompactified flat space, Gauss's law for a charge $q$ and permittivity ${\epsilon }_d$ gives $E_r{A}_{d-1}{r}^{d-1}=q/{\epsilon }_d$. Since $E_r=-d\phi /dr$, the electric potential for $d>2$, normalized to vanish at infinity, isThus the four-dimensional electric potential falls as $1/r$, and the five-dimensional one as $1/r^2$. If a convention counts only spatial dimensions as $D$, replace $d$ by $D$ in the first formula. For $d=2$, the integral instead gives $\phi =-q\log (r/r_0)/(2\pi {\epsilon }_2)$; for $d=1$, it gives $\phi =-q|x|/(2{\epsilon }_1)$ up to a constant. Neither case has a finite zero at infinity.

For a Newtonian gravitational potential ${\Phi }_g$, define the flux-normalized coupling ${\kappa }_d$ by ${\nabla }^2{\Phi }_g={\kappa }_{d}m{\delta }^{(d)}(\mathbf{x})$ and acceleration $-\nabla {\Phi }_g$. Exactly the same flux integration gives the attractive potentialIn $d=3$, ${\kappa }_3=4\pi G_4$ reproduces $-G_{4}m/r$. If $G_D$ is defined by the $D$-dimensional Einstein-Hilbert action $S=(16\pi G_D{)}^{-1}\int \sqrt{-g}R$, the weak static Einstein field equations give ${\kappa }_d=8\pi G_D(D-3)/(D-2)$ for $D>3$. ThereforeIndeed, $R_{00}\simeq {\nabla }^2{\Phi }_g$ and $T_{00}-g_{00}T/(D-2)=(D-3)\rho /(D-2)$ in signature $(-+\cdots +)$. A Poisson-defined Newtonian theory in two spatial dimensions has a logarithmic potential, but pure three-dimensional Einstein gravity has no analogous local Newtonian point-mass force; the relativistic normalization must not be extrapolated to that exceptional dimension.

For an unwarped product with a fixed extra-dimensional metric, normalize the Einstein-Hilbert action by $\frac{1}{2}{M}_{\ast }^{n+2}\int d^{4+n}x\sqrt{-g}R$. Integrating the part containing the four-dimensional curvature over the internal space gives $\frac{1}{2}{M}_{\ast }^{n+2}\mathrm{vol}(S_R^n)\int d^{4}x\sqrt{-g_4}{R}_4$. Thus, using the reduced Planck mass,This is a volume relation for compactification; a curved sphere additionally requires a mechanism supporting and stabilizing its background. It is not by itself a proof that a sphere times flat spacetime solves a vacuum gravitational theory.

With $M_{\ast }=10^3\mathrm{GeV}$, $M_{\mathrm{Pl}}=2.435\times 10^{18}\mathrm{GeV}$ and $1{\mathrm{GeV}}^{-1}=1.97327\times 10^{-16}\mathrm{m}$, the spherical-volume factors $A_1=2\pi$, $A_2=4\pi$, $A_6=16{\pi }^3/15$ giveThe first estimate is astronomical; the second is roughly a tenth of a millimetre. If both Planck scales are instead defined by $S=M_D^{D-2}\int \sqrt{-g}R/(16\pi )$, the same algebraic volume relation uses the unreduced $M_{\mathrm{planck}}=1.221\times 10^{19}\mathrm{GeV}$. Holding that convention's $M_{\ast }=1\mathrm{TeV}$ gives radii larger by $(8\pi {)}^{1/n}$: approximately $4.69\times 10^{12}\mathrm{m}$, $6.80\times 10^{-4}\mathrm{m}$, $2.54\times 10^{-14}\mathrm{m}$. These are different conventions for what is held fixed at one TeV, not different scaling laws.

Finally, compactify the fifth coordinate on a circle of circumference $L=2\pi R$. For a source and observer at the same compact coordinate, the method of images turns a five-dimensional static potential $C/{\rho }^2$ intoOne can derive this image sum without assuming a summation formula. The Fourier transform of $1/(r^2+y^2)$ is $\pi e^{-r|k|}/r$; Poisson summation then gives $\pi [1+2{\sum }_{\ell \ge 1}{e}^{-2\pi \ell r/L}]/(Lr)$, which is the displayed hyperbolic cotangent. ThereforeThe short-distance limit is $C/r^2$; the long-distance limit is the four-dimensional $1/r$ potential. These exponential corrections are the nonzero Kaluza-Klein modes, with masses $2\pi |\ell |/L$. For electrostatics $C=q/(4{\pi }^2{\epsilon }_4)$, so the effective permittivity is ${\epsilon }_3=L{\epsilon }_4$. For a Poisson-normalized gravitational potential, $C=-{\kappa }_{4}m/(4{\pi }^2)$ gives ${\kappa }_3^{\mathrm{eff}}={\kappa }_4/L$. A massless scalar associated with the circle radius also contributes in an unstabilized gravitational compactification; recovering pure four-dimensional Einstein gravity with $G_4=G_5/L$ requires its stabilization or removal. The $1/r$ crossover itself is independent of this tensor-versus-scalar normalization issue.

Use the Fourier transform convention $\hat{w}(E)={\int }_{\mathbb{R}}w(t)e^{itE}dt$ and define the spectral filtering of Hamiltonian terms byThe normalization $\hat{w}(0)=1$ means $\int w=1$. Since $w$ is real and nonnegative, $\hat{w}(-E)=\overline{\hat{w}(E)}$. Thus the stated positive-frequency cutoff also implies $\hat{w}(E)=0$ for $E\le -\Delta$. The integral exists in operator norm, because its integrand has norm at most $w(t)\Vert h_Z\Vert$.

Let $H|{\varphi }_i⟩=E_i|{\varphi }_i⟩$, with $E_i-E_0\ge \Delta$ for $i>0$. In this eigenbasis,Every matrix element coupling the unique ground state to an excited state vanishes, in either direction. HenceMoreover, linearity and conservation of $H$ under its own Heisenberg picture evolution giveEach filtered term is Hermitian, but it need not remain supported on its original set $Z$. The tail assumption supplies stronger long-time control than the integrability used here; no additional tail estimate is needed for the requested identities.

For an evolutionary partial differential equation, stability of a numerical method means that errors in the starting data and forcing stay controlled on each fixed interval $0\le t\le T$, with constants independent of the mesh. Fourier stability analysis is especially effective for a constant-coefficient finite difference method on a uniform infinite or periodic grid: translation invariance makes different Fourier modes evolve independently.

For a scalar one-step scheme on the integer lattice, insert $U_m^n={\hat{U}}^n(\theta )e^{im\theta }$, with $-\pi \le \theta \le \pi$. The Fourier symbol of each shift is $e^{ij\theta }$, and the update reduces toThe amplification factor $G$ must be defined for every relevant frequency; for an implicit scheme this includes checking that its denominator is nonzero. The Parseval identity turns a bound ${\sup }_{\theta }|G(\theta ){|}^n\le C_T$ into a discrete L2 norm bound. Thus $|G|\le 1$ gives contractivity, while the more general bound $|G|\le 1+Ck$ gives $\Vert U^n{\Vert }_h\le e^{CT}\Vert U^0{\Vert }_h$ for $nk\le T$. Conversely, frequencies with amplification uniformly greater than one produce unstable wave packets or periodic Fourier modes under refinement.

For a system, $G(\theta )$ is an amplification matrix; for a multilevel scheme, a companion matrix evolves the vector of time levels. The actual requirement is a uniform bound on matrix powers, not just their spectral radius. A nontrivial Jordan block at a unit-modulus eigenvalue creates polynomial growth in the time index. Even simple eigenvalues inside the unit disk can fail to give a uniform bound if the eigenvector matrices become ill-conditioned as the mesh changes. Uniformly controlled diagonalization of a matrix, or a suitable quadratic energy estimate, resolves this issue. The power boundedness of a two-level Fourier scheme illustrates why root multiplicities and conditioning matter.

For the heat equation, put $\mu =k/h_x^2$. Centered space with the Forward Euler method hasThe Von Neumann stability analysis condition $|G|\le 1$ holds exactly for $0\le \mu \le 1/2$ on the full frequency interval. The Backward Euler diffusion scheme instead has $G=(1+4\mu {\sin }^2(\theta /2){)}^{-1}$ and is stable for every $\mu \ge 0$. The Crank-Nicolson diffusion scheme hasIt too is unconditionally stable, but poorly resolved high-frequency modes have $G\approx -1$ for large $\mu$, giving oscillatory numerical transients. A-stability therefore does not guarantee strong damping; L-stability distinguishes the damping of the Backward Euler method.

For the advection equation $u_t+c{u}_x=0$, let $\nu =ck/h_x$. Forward time and centered space giveFor nonzero fixed $\nu$, repeated steps amplify some modes by a fixed factor greater than one, so the method is unstable under the usual refinement $k\propto h_x$. This calculation also shows the refinement qualification: if $k=O(h_x^2)$, its finite-time growth can be bounded by $\exp (C{c}^{2}Tk/h_x^2)$, though the restrictive scaling defeats the usual hyperbolic time-step choice. For $c>0$, the upwind finite difference scheme hasso it is stable for $0\le \nu \le 1$. The Lax-Wendroff advection scheme instead hasgiving $|\nu |\le 1$. These examples separate the effects of the spatial stencil, temporal approximation and Courant number; a higher order of a numerical method by itself does not establish stability.

A periodic boundary condition is ideal for this analysis. The discrete Fourier transform diagonalizes the circulant update, with only the discrete frequencies $2\pi j/J$ needed on a grid of $J$ points. A bound on all $\theta$ is a convenient guarantee over every such grid. On the whole line, the Fourier transform uses a continuous frequency interval and the Parseval identity proves the corresponding square-summable-data estimate.

Homogeneous Dirichlet boundary conditions do not admit arbitrary complex exponential Fourier modes. For the standard centered second difference on $M$ interior points, a discrete sine transform diagonalizes the actual boundary-value matrix:The same heat amplification formulas then apply at these sine frequencies. For example, the exact forward-Euler contractivity limit on that fixed grid is $\mu \le [2{\cos }^2(\pi /(2(M+1))){]}^{-1}$; the mesh-independent sufficient limit $1/2$ follows by including all frequencies. Homogeneous Neumann boundary conditions can similarly permit a discrete cosine transform, provided the endpoint discretization and its weighted inner product are chosen consistently. The constant mode is then present, reflecting conservation of the heat equation's spatial mean. For inhomogeneous boundary conditions, subtract a suitable lifting of the prescribed boundary data and estimate the induced source using the homogeneous evolution's bound and a Duhamel principle estimate. Bounds for the lifting and source must themselves be uniform in the mesh.

General boundary conditions require separate boundary stability of a finite-difference method. For an advection equation, incoming data are prescribed at the inflow boundary, while an outflow closure must respect the outgoing characteristics. An interior periodic Fourier symbol cannot detect a growing mode confined near a boundary. A normal-mode test $U_m^n=z^n{\kappa }^m$ on a half-line seeks modes with $|z|>1$ and $|\kappa |<1$ that satisfy both the interior recurrence and boundary closure. Their existence proves instability, and uniform control requires more than merely excluding isolated growing roots; the Uniform Kreiss--Lopatinskii condition addresses boundary resolvent bounds. Alternatively, a direct energy method using summation by parts can include the boundary terms and establish the required estimate.

Variable coefficients and nonuniform meshes usually destroy exact Fourier transform diagonalization. Frozen-coefficient Fourier stability analysis is then a useful diagnostic, but it is not automatically a proof for the full variable-coefficient boundary problem. The mesh-uniform energy method in question 5 is an example of a direct proof. Also, an L2 norm proof is not automatically a mesh-uniform maximum-norm proof: finite-dimensional norm-equivalence constants may grow with the number of grid points.

Finally, stability measures error propagation; consistency of a numerical method measures the defect of inserting the exact solution. For a well-posed linear initial-value problem in the chosen norm, the Lax equivalence theorem says that a consistent approximation is convergent exactly when it is stable. Boundary discretization and starting data must be included in that assertion. A Fourier calculation proves convergence only after consistency, well-posedness and the actual boundary treatment have also been checked.

Use the unnormalized Fourier transform in the question and the Plancherel theorem, with inverse factor $1/(2\pi )$. For a general $L^2$ scaling function, the transform is interpreted in the $L^2$ sense; the printed integral need not be absolutely convergent. All frequency identities are almost-everywhere statements.

Changing variables in the Fourier transform givesThus taking the transform of the scaling refinement equation gives $f(\xi )=m(\xi /2)f(\xi /2)$, orConversely the same calculation and injectivity of the Fourier transform recover refinement, with convergence understood in $L^2$.

To justify both the orthogonality assertion and this convergence precisely, put $P(t)={\sum }_{k\in \mathbb{Z}}|f(t+2\pi k){|}^2$. It belongs to $L^1[-\pi ,\pi ]$ by monotone integration. The Plancherel theorem givesHence orthonormality of all integer translates is equivalent, by uniqueness of Fourier coefficients in L1, to the periodized-energy conditionWhen this holds, the squared norm of ${\sum }_n{a}_n\varphi (2\cdot -n)$ is $\frac{1}{2}{\sum }_n|a_n{|}^2$, so square-summable coefficient series converge in $L^2$. In the converse direction the refinement identity and $P=1$ give, by splitting the periodization of $|f(2t){|}^2$ into even and odd translates,Thus $m$ is bounded and its Fourier coefficients $a_n/2$ are square summable. If $m_N$ are its Fourier partial sum, thenThis verifies the transformed refinement series converges to $f$, completing the equivalence of the two pairs of conditions without imposing an unnecessary $L^1$ assumption on $\varphi$.

The half-open intervals $[-\pi +2\pi k,\pi +2\pi k)$ tile the line, so exactly one term contributes to ${\sum }_k|f(t+2\pi k){|}^2$. It equals one almost everywhere.

Choose the Shannon scaling mask, a $2\pi$-periodic low-pass filter of a multiresolution analysis which equals one on $[-\pi /2,\pi /2)$ and zero on the rest of $[-\pi ,\pi )$. On the support of $f$ its product with $f(t)$ equals $f(2t)$; outside that support both sides vanish. Its Fourier coefficients giveso it also has the required symbol representation. The inverse Fourier transform gives the Shannon scaling functionThis sinc function has $L^2$ norm one. It illustrates why the Fourier transform convention in part B must allow $L^2$ transforms: the Shannon scaling function is not absolutely integrable on the line.

Use the half-line Fourier transform and finite-time boundary transformsThe half-line Fourier transform is analytic for $\mathrm{Im}k<0$ under spatial decay. All spectral integrals below have their usual oscillatory, or vanishing Gaussian damping, interpretation until absolute convergence is established.

The free Schrodinger equation has the local divergence identityIntegrating in $x,t$ gives the global relation for the half-line free Schrodinger equationLet $D_{+}=\{\mathrm{Re}k>0,\mathrm{Im}k>0\}$. Orient its boundary from $i\infty$ down to $0$, and then from $0$ to $+\infty$, so that $D_{+}$ lies on the left. Fourier inversion followed by contour integration in the second quadrant yieldsThis is a complex spectral representation involving the initial trace and both boundary traces. The contour deformation works because each boundary-time integrand contains $e^{-i{k}^2(t-s)}$ with $s\le t$, which decays in the second quadrant, as well as $e^{ikx}$ for $x>0$. This fixes both the quadrant and the orientation; changing either without changing the signs would produce an incorrect representation.

We place the definitions and the unheaded preliminary requests here before addressing the first labelled property. The Schwartz space consists of smooth functions with finite seminormsfor every pair of multi-indices. Convergence means convergence in each seminorm. The tempered distribution space ${\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ is its continuous dual space, with weak convergence tested against every Schwartz function. A continuous functional satisfies a bound by finitely many of these seminorms, equivalently by one sufficiently large weighted derivative seminorm.

Use the Fourier transform conventionDifferentiation under the integral and integration by parts express ${\xi }^{\alpha }{\partial }_{\xi }^{\beta }\hat{f}$ as a constant of modulus one times the Fourier transform of ${\partial }_x^{\alpha }(x^{\beta }f)$. Its supremum is bounded by the $L^1$ norm of that function. For $s>n$, this norm is at most a constant times finitely many Schwartz seminorms, using the integrable weight $(1+|x|{)}^{-s}$. Thus $\mathcal{F}:\mathcal{S}\to \mathcal{S}$ is continuous.

For completeness, Fourier inversion follows by inserting $e^{-\epsilon |\xi {|}^2}$ in the inverse integral and using Fubini's theorem. The result is convolution with the Gaussian approximate identityIt tends to $f$, while integrability of $\hat{f}$ allows the damping factor to be removed by dominated convergence theorem. Consequently ${\mathcal{F}}^{2}f=(2\pi {)}^{n}f(-\cdot )$. Reflection preserves every Schwartz seminorm, so the inverse transform is continuous as well. This proves the Fourier transform isomorphism of the Schwartz space.

Define the Fourier transform of a tempered distribution by transposition,The Schwartz-space continuity just proved makes this a tempered distribution. Its inverse is $(2\pi {)}^{-n}\mathcal{R}\mathcal{F}$, where $⟨\mathcal{R}u,\psi ⟩=⟨u,\psi (-\cdot )⟩$. These maps are continuous for weak convergence, since each pairing is a pairing with a fixed transformed test. They are also continuous for the strong dual topology, because the Schwartz-space maps take bounded sets to bounded sets.

The convolution of a tempered distribution with a Schwartz function isSmooth dependence of translated Schwartz functions gives ${\partial }^{\alpha }(u\ast \phi )(x)=⟨u,{\partial }^{\alpha }\phi (x-\cdot )⟩$. The finite-seminorm estimate and $1+|y|\le (1+|x|)(1+|x-y|)$ show that each derivative has at most polynomial growth. In particular, $u\ast \phi$ is a smooth function defining a tempered distribution. It need not itself be a Schwartz function; for example $1\ast \phi =\int \phi$.

Writing $\check{\phi }(y)=\phi (-y)$, its distributional pairing is $⟨u\ast \phi ,\psi ⟩=⟨u,\check{\phi }\ast \psi ⟩$. The inner convolution is a Schwartz function, and this identity follows by integration in the Schwartz topology, justified by the weighted seminorm estimates. A direct Fubini's theorem calculation gives $\check{\phi }\ast \hat{\psi }=\mathcal{F}(\hat{\phi }\psi )$. HenceMultiplication is well defined because multiplication by $\hat{\phi }$ acts continuously on $\mathcal{S}$.

Now write the Hilbert transform as convolution with $K={\pi }^{-1}\mathrm{pv}(1/x)$, the principal-value reciprocal distribution. The given Heaviside function transform, together with ${\mathcal{F}}^{2}H=2\pi H(-\cdot )$, yieldsThe value of the sign function at zero is irrelevant to its regular distribution. Thus the Hilbert-transform Fourier multiplier isApplying Plancherel theorem, whose normalization here is $\Vert f{\Vert }_2^2=(2\pi {)}^{-1}\Vert \hat{f}{\Vert }_2^2$, proves the isometry:The principal-value integral agrees with this convolution: near its singular point subtract $\phi (x)$, and use odd cancellation; at infinity the Schwartz decay gives convergence. The Hilbert transform has domain $\mathcal{S}$, but generally does not take values in $\mathcal{S}$, as the tail in part (c) demonstrates.

The Hilbert-transform Fourier multiplier is bounded, and ${\xi }^r\hat{\phi }(\xi )$ is integrable for every nonnegative integer $r$. Therefore its inverse Fourier transform can be differentiated under the integral arbitrarily many times:These derivatives are continuous by dominated convergence theorem. Since $\widehat{{\phi }^{(r)}}=(i\xi {)}^r\hat{\phi }$, the Hilbert transform is smooth and commutes with differentiation:The Fourier proof avoids differentiating a singular kernel without preserving its Cauchy principal value prescription.

Let $f=\hat{\phi }$, a Schwartz function. Splitting the inverse Fourier transform at the jump in the Hilbert-transform Fourier multiplier givesFor $x\ne 0$, integration by parts on each half-line showsBoth restricted derivatives are in $L^1$. The Riemann-Lebesgue lemma makes the bracket tend to zero as $|x|\to \infty$. Hence the two-sided tail isHere $\hat{\phi }(0)=\int \phi$. The large-distance tail of the Hilbert transform thus depends on the zeroth moment of the input. If that moment is nonzero, the $1/x$ tail proves that the output is not a Schwartz function, despite being smooth and square-integrable.

Consider the operators on all of $C_0(X)$ or $L^2(\mu )$, extending the displayed positive-function formula linearly. Here $C_0(X)$ is the space of continuous functions vanishing at infinity. The multiplier has modulus at most one, so $\Vert P_{t}f\Vert \le \Vert f\Vert$, and multiplication of exponentials gives $P_{s+t}=P_s{P}_t$ and $P_0=I$. For strong continuity on $C_0(X)$, choose a compact $K$ outside which $|f|$ is small. On $K$ the continuous $k$ is bounded, making $e^{-tk}\to 1$ uniform; outside $K$, $|(e^{-tk}-1)f|\le |f|$. For $L^2(\mu )$, use pointwise convergence and the dominated convergence theorem with bound $|(e^{-tk}-1)f{|}^2\le |f{|}^2$. Thus both are contraction semigroups, even when $k$ is unbounded.

The generator of a multiplication semigroup isIndeed pointwise difference quotients converge to $-kf$. A norm limit therefore forces the displayed domain condition, using an almost-everywhere convergent subsequence for the $L^2$ case. Conversely $|(e^{-tk}-1)/t|\le k$, so the dominated convergence theorem proves the $L^2$ limit when $kf\in L^2$. In $C_0(X)$ write the error as $kf$ times $1-(1-e^{-tk})/(tk)$, with the value at $k=0$ supplied by continuity. That second factor tends uniformly to zero where $k$ is bounded and has modulus at most one; the compact-set and small-tail argument applied to $kf$ proves uniform convergence. To prove directly that $L$ is a closed operator, take $f_n\to f$ and $L{f}_n\to g$. Pointwise limits in $C_0$, or a common almost-everywhere convergent subsequence in $L^2$, give $g=-kf$. Thus $f$ is in the appropriate generator domain and $Lf=g$. In $L^2$ the domain is dense because $f{1}_{\{k\le n\}}\to f$; in $C_0$ compactly supported continuous functions are a dense subspace contained in the domain.

A real multiplication operator on its maximal $L^2$ domain is an unbounded self-adjoint operator. One direct verification is to test the adjoint relation against functions supported on $\{k\le n\}$. If $h$ is in the adjoint operator domain with representing vector $g$, those tests force $g=-kh$ on each such set, hence $kh\in L^2$. The converse follows by integration. On the sigma-finite measure spaces used below, the spectrum of a real multiplication operator is its essential range; spectral values are detected by unit vectors supported where the multiplier is arbitrarily close to that value, while outside the essential range its reciprocal is a bounded resolvent multiplier.

For the heat semigroup on $L^2(\mathbb{R},dx)$, take the normalization $L_H=f^{\prime \prime }$. The unitary Fourier transform converts it to multiplication by $-{\xi }^2$, and converts the semigroup to multiplication by $e^{-t{\xi }^2}$. ConsequentlyThe Sobolev space domain is exactly $\{f:{\xi }^2\hat{f}\in L^2\}$. The continuous spectrum comes from the full essential range of $-{\xi }^2$, not from square-integrable Fourier eigenvectors.

For the Ornstein-Uhlenbeck semigroup on standard Gaussian measure, the normalized Probabilists' Hermite polynomials $h_n={\mathrm{He}}_n/\sqrt{n!}$ form an orthonormal basis. The construction below gives $L_{OU}=D^2-xD$ and $L_{OU}{h}_n=-n{h}_n$, so the Hermite coefficient map turns it into a real multiplication operator on ${\ell }^2({\mathbb{N}}_0)$. ThereforeFinally, the heat semigroup on the whole real line has no positive spectral gap, whereas the standard Ornstein-Uhlenbeck semigroup has gap one. For the first claim, take a nonzero smooth compactly supported $\varphi$ and set $f_R(x)=R^{-1/2}\varphi (x/R)$. Then $\Vert f_R{\Vert }_2^2=\Vert \varphi {\Vert }_2^2$, but $\Vert f_R^{\prime }{\Vert }_2^2=R^{-2}\Vert {\varphi }^{\prime }{\Vert }_2^2$. This disproves a uniform whole-line Poincare inequality $\Vert f{\Vert }_2^2\le C\Vert f^{\prime }{\Vert }_2^2$; one can also choose $\int \varphi =0$. Lebesgue measure here is infinite, so it has no normalized probability mean to subtract. For Gaussian measure, the Hermite expansion instead gives the sharp Gaussian Poincaré inequalityEquality holds for affine functions. The identity on the right is interpreted on the energy-form domain, which is larger than the full operator domain.

PutIn free space, the parabolic wave equation is $E_x=i{E}_{zz}/(2k)$. Under the Fourier transform convention $\hat{E}(p)=\int E(z)e^{-ipz}dz$, it becomesThe Gaussian integral is legitimate because $\mathrm{Re}A=1/D^2>0$. Invert the transform after multiplying by $e^{-i{p}^{2}x/(2k)}$. A second Gaussian integral, or equivalently one-dimensional transverse Fresnel propagation, givesChoose the square-root branch continuously from $q(0{)}^{-1/2}=1$; for real $x\ge 0$ there is no zero of $q$. This gives the correct incident field at $x=0$, and direct differentiation verifies the free parabolic wave equation.

For clarity, the squared envelope magnitude isThe Gaussian beam with one transverse coordinate remains Gaussian, with one-transverse-coordinate amplitude factor $q^{-1/2}$, not the $q^{-1}$ of a beam with two transverse coordinates. The negative initial quadratic phase produces focusing for $F>0$; diffraction prevents a singularity at $x=F$. These expressions describe the paraxial approximation to free propagation, rather than an exact unrestricted Helmholtz equation beam.

Let the medium's autocorrelation function of a random field and power spectrum use the explicit conventionSince $n^2-1=2\mu W+{\mu }^2{W}^2$, only the linear term in $\mu$ is needed to obtain the cross-section through second order:Squaring and taking the expectation givesThis is the finite-volume formula in the Born approximation for scalar wave scattering; the zero mean of $W$ does not make its squared scattered amplitude vanish. For a jointly centered Gaussian random field, the third-order contribution is zero, so the next perturbative contribution is fourth order when that expansion is valid.

Introduce the overlap volume $H_V(\mathbf{s})=|V\cap (V-\mathbf{s})|$. Changing variables rewrites the double integral asWhen the linear dimensions of a regular scattering volume are large compared with the correlation length, and $C$ is integrable, $H_V(\mathbf{s})\simeq |V|$ over the significant covariance range. This gives the finite-volume random scattering spectrum resultThere are two distinct approximations here: the weak-scattering expansion in $\mu$ and the negligible-boundary-overlap approximation in the volume-to-correlation-scale ratio. If $S$ instead denotes the power spectrum of the index fluctuation $n-1=\mu W$, then $S={\mu }^2{S}_W$ and the external factor ${\mu }^2$ is absorbed into $S$.

Statistical homogeneity alone allows an anisotropic spectrum, so its argument is generally the vector ${\mathbf{k}}_s$. If statistical isotropy is additionally assumed, $S_W$ depends only onThe scalar notation $S(k_s)$ is valid in that isotropic case. For the alternative convention $S_{\mathrm{norm}}=(2\pi {)}^{-3}{S}_W$, the bulk formula is $2\pi {\mu }^2{k}^4|V|S_{\mathrm{norm}}({\mathbf{k}}_s)$. Changing the sign of the scattering vector does not change the spectrum of a real stationary scalar field; changing the Fourier transform normalization does change the displayed $2\pi$ factors. Both conventions have therefore been stated explicitly.

Use the Fourier transform convention $\hat{p}(k,y)={\int }_{\mathbb{R}}{e}^{ikx}p(x,y)dx$. Let $\gamma (k{)}^2=k^2-k_0^2$, choosing $\mathrm{Re}\gamma >0$ on the real contour with $\mathrm{Im}\omega <0$. For time dependence $e^{i\omega t}$, this is the decaying continuation of the outgoing Sommerfeld radiation condition.

The transformed Helmholtz equation has upper and lower solutions $P(k)e^{-\gamma y}$ and $Q(k)e^{\gamma y}$. The two kinematic traces give $-\gamma P={\rho }_0{\omega }^2\hat{\eta }=\gamma Q$, henceThe elastic membrane equation gives $[\hat{p}]=(m{\omega }^2-T{k}^2)\hat{\eta }$. Therefore the acoustic wave on a tensioned massive membrane has dispersion relationEquivalently, $T{k}^2={\omega }^2(m+2{\rho }_0/\gamma )$. The fluid on both sides supplies a positive added mass of an evanescent fluid layer, which is a useful independent check on the sign. In particular, the incident pressure isThe symbol $m$ here denotes elastic membrane mass per area, rather than the fluctuating Mach number used in Question 1.