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Partition the range into planes $x_j=x_0+j\Delta x$. At each step, evaluate the refractive-index phase screen at $x_j$ or the midpoint, multiply the current envelope by $e^{\Delta x{S}_j}$, transform in $z$, multiply each transverse Fourier component by the diffraction phase $e^{-i{k}_z^2\Delta x/(2k_0)}$, and apply the inverse Fourier transform. Repeating this split-step Fourier method $m$ times produces $E(x_m,z)$ from $E(x_0,z)$. The step size must continue to resolve both longitudinal medium variation and the Lie-Trotter splitting error.

Withthe parabolic wave equation is $E_x=(D+S)E$. Freeze $S$ at $x_0$, or preferably at the step midpoint. Over a short distance $\Delta x$, Lie-Trotter splitting givesThe reversed ordering has the same first-order accuracy, while symmetric half-steps in $S$ give the more accurate Strang form.

The commutator can be displayed explicitly. If $q=n^2-1$, thenThe splitting assumption requires $\Delta x^2[D,S]E$ to be small relative to $E$. It is favored by a short range step, a transversely smooth refractive index, and a field without unresolved large transverse wavenumbers. Freezing $S$ also requires $\Delta x{S}_x$ to be small. These conditions supplement the one-way and paraxial approximation already used in part i.

The phase-screen substep is pointwise:DefineThen solve the free-diffraction initial-value problemto $x_0+\Delta x$. In transverse Fourier transform variables, this substep is simplyThis is the split-step Fourier method.

The Fluctuation-dissipation theorem in this convention readsIn the classical limit $|\omega |\ll T$, this becomesThe inverse temporal Fourier transform is therefore

Write $D={\rho }_s/s$, so ${\omega }_k=D{k}^2$, and take ${\Gamma }_k=\gamma |k|$. The small-wavenumber static structure factor behaves asThe spatial Fourier transform of $|k{|}^{-1}$ in $d$ dimensions scales as $|x{|}^{1-d}$. Thusand in three dimensions $C(x)\sim |x{|}^{-2}$. Since this two-point correlation function tends to zero rather than to a nonzero constant at large separation, it is incompatible with true long-range ferromagnetic order and hence with spontaneous symmetry breaking at this higher temperature.

Position-space particle-hole symmetry complex-conjugates plane waves, so the Fourier transform sends $k$ to $-k$. Fourier transforming the stated relation therefore givesIf $H_{\mathrm{BdG}}(k)u_k={\epsilon }_k{u}_k$, complex conjugation and multiplication by ${\Sigma }_1$ yieldThus the Particle-hole symmetry of a Bogoliubov--de Gennes Hamiltonian imposesor, band by band after a suitable relabelling, ${\epsilon }_{-k}=-{\epsilon }_k$.

Use the Fourier transform conventionThe zero mode vanishes because the compositional order parameter has zero spatial average. Parseval identity and the Fourier transform of a derivative turn the quadratic part of the dimensionless free energy intoHere ${\sum }_{\mathbf{q}}^{+}$ is the positive-wavevector sum for a real field. Each independent complex amplitude has density proportional to $e^{-G(q)|{\varphi }_{\mathbf{q}}{|}^2}$, so its elementary Gaussian integral gives the static structure factorwhenever $G(q)>0$.

The stationary points of the Brazovskii model kernel obeyBecause $\kappa <0<\gamma$, the nonzero minimum is the nonzero-wavevector soft-mode sphereAt this wavevector,The first Gaussian field theory divergence therefore occurs at

After the sudden parameter change, the functional derivative isEach Cartesian component of each Fourier mode consequently obeysThis is an Ornstein-Uhlenbeck process. The integrating-factor method givesBecause both coefficients are positive, every mode has positive decay rate.

Use normalized convolution and Fourier coefficients, and define the large spectrumBy the Parseval identity and $0\le f\le 1$,so $|\Gamma |\le {ϵ}^{-2}{\delta }^{-1}$.

The convolution theorem givesIf $y\in B(\Gamma ,ϵ)$, the part over $\Gamma$ is at mostbecause $|\hat{f}(\gamma )|\le \delta$. On the complementary spectrum, $|\hat{f}(\gamma )|<ϵ\delta$ and $|\gamma (y)-1|\le 2$, so the contribution is less than $2ϵ{\delta }^2$. The required difference is therefore less than $3ϵ{\delta }^2$.

Let $U$ be uniform on $[0,2\pi ]$, and independently let $W$ have densityThe supplied Fourier transform identity, after the change of Fourier convention, givesAlso ${\mathbb{E}}_U[\cos (a+U)\cos (b+U)]=\frac{1}{2}\cos (a-b)$. Thereforeso one may take $c=1$.

For this metric, $\sqrt{-g}=a^2$, while $g^{\eta \eta }=-a^{-2}$, $g^{zz}=a^{-2}$ and $g^{ij}={\delta }^{ij}$. The massless Klein-Gordon equation is consequentlyInsert the spatial Fourier transform given in the question. Each wavenumber mode then satisfiesThus every field mode is a simple harmonic oscillator with a time-dependent angular frequency.

Extend $f$ by zero outside $[0,1]$. Since the Fourier transform of $e^{-|x|}$ is $2/(1+{\xi }^2)>0$,for $f\ne 0$. Thus every eigenvalue is positive. From the relation in part (b),The transcendental equation has its successive roots in intervals separated by the poles and zeros of $\tan \omega$, so ${\omega }_n$ grows linearly with $n$. Hence

For a compact convex set $K\subset {\mathbb{R}}^n$, letThe Paley–Wiener–Schwartz theorem says that if $u$ is a compactly supported distribution with support in $K$, its Fourier--Laplace transformis entire and, for some $C,N$,Conversely, every entire function satisfying such an estimate is the transform of a distribution supported in $K$.

For the forward direction, compact support lets $u$ act on the exponential after insertion of a cutoff equal to one near $K$. Differentiation in $z$ may be passed under the pairing, proving entire analyticity. The finite-order estimate for $u$ bounds derivatives of the exponential on $K$ by a polynomial in $|z|$ times $e^{H_K(\mathrm{Im}z)}$.

Conversely, restrict the entire function $F$ to ${\mathbb{R}}^n$. Its polynomial growth defines a tempered distribution $u$ by inverse Fourier transform. If a test function is supported outside $K$, separate its compact support from $K$ by a real vector $\eta$. Shifting the Fourier inversion contour from ${\mathbb{R}}^n$ to ${\mathbb{R}}^n+it\eta$ is allowed by entire analyticity. The exponential gained from the test function beats the bound $e^{t{H}_K(\eta )}$ as $t\to \infty$, so the pairing vanishes. Hence $\mathrm{supp}u\subset K$, completing the converse.

LetApplying the parabolic wave equation to each factor givesIf $C(s)=⟨w(z)w(z+s)⟩$, Gaussian averaging at the screen givesStationarity makes this a function only of $s=z_1-z_2$. Since ${\partial }_{z_1}^2+{\partial }_{z_2}^2=2{\partial }_s^2$ on such functions,Writing ${\hat{M}}_0(q)$ for the Fourier transform of the screen value, the solution at arbitrary range is

The spectral characterization of the Reproducing-kernel Hilbert space of a stationary Gaussian process says that its elements are precisely the functions whose Fourier transforms satisfywith $c$ determined only by the Fourier transform convention. Here $\hat{K}(u)=(1+u^2{)}^{-1}$, soThis is an equivalent norm for the Sobolev space $H^1(\mathbb{R})$, and hence the RKHS equals $H^1(\mathbb{R})$ as a set.

Since $\hat{K}$ is integrable, $K$ is continuous and the process has a jointly measurable separable version. For every finite Borel measure $\nu$ on $\mathbb{R}$, Tonelli theorem givesThus $X\in L^2(\mathbb{R},\nu )$ almost surely and is a Borel random variable there. Every continuous linear functional of $X$ is a centered normal random variable: approximate its $L^2(\nu )$ integral by finite linear combinations of process values and pass to the $L^2$ limit. Therefore the induced law is a Gaussian Borel measure on $L^2(\mathbb{R},\nu )$.

After a radial Fourier transform, Poisson's equation away from the sheet becomesThe decaying, reflection-symmetric solution is $A{e}^{-|k||z|}$. Integrating Poisson's equation across $z=0$ gives the derivative jumpso $-2|k|A=4\pi G\tilde{\Sigma }$. Thus the midplane potential is

A parametrix for $P(D)$ is a distribution $E$ for whichwith $r\in C^{\infty }$. Choose a smooth cutoff $\chi$ that vanishes on a large ball containing every real zero of $P$ and equals one outside a slightly larger ball. Ellipticity makesa symbol of order $-N$. For $E={\mathcal{F}}^{-1}q$,Since $1-\chi$ is smooth and compactly supported, its inverse Fourier transform is smooth. Thus $E$ is a parametrix.

For a Schwartz function on the real numbers, its periodization is a smooth periodic function of period one. Rapid decay gives uniform convergence on $[0,1]$ of the series and each differentiated series. Under the Fourier transform convention $\hat{f}(\xi )=\int f(x)e^{-2\pi ix\xi }dx$, its $m$th Fourier coefficient is $\hat{f}(m)$: integrate termwise over one period and combine the translated intervals. The rapidly decreasing coefficients give a Fourier series with uniform convergence. Evaluation at zero proves the Poisson summation formula.

A harmonic line force localized at one point of a two-dimensional sheet gives $\hat{\eta }=F/D$ and $\Delta =\gamma D$ in the Fourier transform. The outgoing upper pressure is the displayed amplitude. Dividing by $c_0^2$ converts it to acoustic density perturbation. The method of steepest descent gives cylindrical sound from the saddle point, while crossed zeros of the dispersion relation give guided or leaky modal residues. Retaining $k_0=\omega /c_0$ is essential for dimensionally consistent spreading and prefactors.

For a sufficiently rapidly decreasing function $f$, the Poisson summation formula relates its samples to those of its Fourier transform:under the convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-ix\xi }dx$.

A forcing pole can be split between analytic half-planes by subtracting a factor's value at the pole. For instance,The second quotient has a removable pole and belongs to the plus expression; the first keeps the prescribed minus-side forcing pole. After Wiener-Hopf factorization, analytic continuation identifies an entire remainder, whose value is fixed by growth and edge conditions. The pole prescription must be stated together with the Fourier transform convention.