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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.

Transform pair. In the paper's symmetric normalization, the inverse Fourier transform isSplit the transform of $e^{-c|x|}$ at zero, with $c>0$:The inverse formula applied to this result givesBecause the kernel is even, replacing the transform variable by $x$ and the argument by $k$ yieldsThe factors of $\sqrt{2\pi }$ depend on the normalization and will also enter the convolution theorem.

Convolution theorem. For integrable functions define the convolutionSubstitute this into the Fourier transform, interchange the two absolutely integrable factors using Fubini's theorem, and set $z=x-y$:Therefore

Integral equation. Put $f_a(x)=1/(x^2+a^2)$. Since this is even, the left side is $f_a\ast g$. The transformed equation issoThe assumption $b>a$ makes the new scale $d=b-a$ positive. Inverting with the previously established transform pair givesThe Cauchy convolution semigroup gives a quick verification. With $K_c(x)=c/[\pi (x^2+c^2)]$, the formula is $g=(a/b)K_{b-a}$ and $f_a=(\pi /a)K_a$. Hence $f_a\ast g=(\pi /b)K_b=1/(x^2+b^2)$, as required. The solution is unique among integrable functions because the multiplier ${\tilde{f}}_a(k)$ never vanishes.

Laplace transform. Use $Y(p)={\int }_0^{\infty }{e}^{-pt}y(t)dt$. The initial value is zero and the Dirac delta contributes $e^{-p}$, so the transformed equation is $pY+Y=e^{-p}$. Therefore

Bromwich inversion. The Bromwich inversion formula iswith the vertical Bromwich contour traversed upwards and its symmetric improper limit understood. For $t>1$, close to the left, where the exponential decays; the enclosed simple pole at $p=-1$ has residue $e^{-(t-1)}$. For $t<1$, close to the right, which contains no pole. Thuswhere $\Theta$ is the Heaviside step function. This agrees with the Laplace transform time-shift rule. At $t=1$, inversion returns the average $1/2$ of the two one-sided values. The value assigned at that single point does not affect the differential equation in the distributional sense.

Fourier transform. State the convention explicitly:The Fourier transform of the distributional derivative is $ik\tilde{y}(k)$, including the jump at $t=1$. The Dirac delta has transform $e^{-ik}$, giving

Inverse Fourier integral. Substitute into the inverse Fourier transform:The only pole is at $k=i$. For $t>1$, close in the upper half-plane. Jordan's lemma removes the large arc and the residue theorem gives $i{\mathrm{Res}}_{k=i}=i[e^{-(t-1)}/i]=e^{-(t-1)}$. For $t<1$, close in the lower half-plane, where there are no poles, giving zero. This reproducesIntegrating the differential equation across the impulse checks the unit jump, $y(1^{+})-y(1^{-})=1$. Away from the impulse the solution obeys $y^{\prime }=-y$, and it satisfies the stipulated zero past.

Transform convention and derivatives. Use the Fourier transform pairFor sufficiently decaying $f,f^{\prime }$ with integrable second derivative, two applications of integration by parts giveThe identity also holds for tempered distributions, which is the setting needed for the Dirac delta function boundary data.

Upper half-plane solution. Take the Fourier transform in $x$ only, writing $\Phi (k,y)=\tilde{\varphi }(k,y)$. Laplace's equation becomes ${\Phi }_{yy}-k^2\Phi =0$. For $k\ne 0$, discard the exponentially growing solution and use $\Phi (k,0)=1$, the Fourier transform of the Dirac delta function. This gives $\Phi (k,y)=e^{-|k|y}$. The value at $k=0$ can be set to its continuous limit one. Pointwise decay of $\varphi$ as $y\to \infty$ does not require this individual transformed value to decay: the total integral in $x$ remains one while the profile spreads.

The inverse Fourier transform now yields, for $y>0$,Evaluate the last integral as the real part of ${\int }_0^{\infty }{e}^{-(y-ix)k}dk=(y-ix{)}^{-1}$:This is the Poisson kernel for the upper half-plane. Its boundary condition is a distributional convergence, rather than an ordinary value at $(0,0)$: for a smooth compactly supported test function $g$,The dominated convergence theorem applies because $g$ is bounded and $(1+t^2{)}^{-1}$ is integrable. Also $0\le \varphi (x,y)\le 1/(\pi y)$, proving the required decay as $y\to \infty$.

Direct verification. Differentiating the rational expression givesTheir sum is zero throughout $y>0$, so the solution is a harmonic function there.

Use the Fourier transform convention $\hat{f}(k)={\int }_{\mathbb{R}}f(t)e^{-ikt}dt$, with inverse factor $1/(2\pi )$. For functions in L1 space, their convolution is $h(t)={\int }_{\mathbb{R}}f(s)g(t-s)ds$. Fubini's theorem applies because the absolute double integral is $\Vert f{\Vert }_1\Vert g{\Vert }_1$. Substituting $r=t-s$ therefore proves the convolution theorem:For the specified rectangular pulse, the support condition is $|t|<1$. Its Fourier transform isThe convolution $f\ast f$ is the length of the overlap of $(-1,1)$ and $(t-1,t+1)$. Thus it is the triangular pulseThe transform's value at zero is four, given by its continuous extension. In this convention, the Parseval identity for square-integrable functions isFor the rectangular pulse the left side is two, giving the sinc-squared integralFor the triangular pulse, $\int |f\ast f{|}^{2}dt=2{\int }_0^2(2-t{)}^{2}dt=16/3$. Apply the same Parseval identity with transform $4{\sin }^{2}k/k^2$ to obtain the sinc fourth-power integralBoth integrands have removable singularities at zero and integrable tails.

Use the Fourier transform conventionThese hold directly under suitable integrability and inversion hypotheses, and extend to square-integrable functions through the Plancherel theorem. At a jump, ordinary Fourier inversion uses the mean of the one-sided limits under the usual piecewise smooth hypotheses.

Use the conventionThese are the Fourier transform and inverse Fourier transform for suitably integrable data. Assuming the required ordinary transforms exist and the lower derivatives vanish at both ends, integration by parts gives $\widehat{g^{\prime }}=[g{e}^{-ikx}{]}_{-\infty }^{\infty }+ik\hat{g}=ik\hat{g}$. Repeating this argument yieldsThe decay assumptions are used to discard the boundary terms at each step.

For positive diffusivity $\alpha$, transforming the heat equation gives ${\partial }_t\hat{f}=-\alpha k^2\hat{f}$. Thus $\hat{f}(k,t)=e^{-\alpha k^{2}t}{\hat{f}}_0(k)$, where $f_0(x)=f(x,0)$. The convolution theorem and the Gaussian heat kernel giveFor $f_0(x)=e^{-x^2}$, the Fourier transform of a Gaussian is $\sqrt{\pi }{e}^{-k^2/4}$. Multiplying by the heat factor and inverting gives the Gaussian initial data for the heat equation solutionIt recovers the initial profile as $t\downarrow 0$ and preserves its spatial integral $\sqrt{\pi }$.

Take the spatial Fourier transform with the same convention. The partial differential equation becomes $i{\partial }_t\hat{f}-k^2\hat{f}=0$, so $\hat{f}(k,t)=e^{-i{k}^{2}t}{\hat{f}}_0(k)$. Inverting and substituting the transform of the initial function yields the convolutionComplete the square, $iks-it{k}^2=i{s}^2/(4t)-it[k-s/(2t){]}^2$, and put $\ell =\sqrt{t}[k-s/(2t)]$. This givesThe oscillatory Gaussian is a Fresnel integral, understood as the limit with a positive Gaussian damping factor. That prescription also justifies the transform manipulations, first for rapidly decreasing smooth data and then by L2 space extension. The resulting kernel is the free-particle propagator in units $\hslash =1$, $m=1/2$; its phase and normalization agree with the multiplier $e^{-i{k}^{2}t}$.

Extend $f$ evenly to $F(x)=f(|x|)$ on the whole real line, and use the Fourier transform convention $\hat{F}(k)={\int }_{\mathbb{R}}F(x)e^{-ikx}dx$. Its odd sine contribution vanishes, giving $\hat{F}(k)=2{\int }_0^{\infty }f(x)\cos (kx)dx={\hat{f}}_c(k)$ for $k>0$; the transform is also even. The Fourier inversion theorem therefore gives, for $x>0$,This is the inverse Fourier cosine transform with the factor-of-two forward convention used here.

The even extension is the triangular function $f(x)=(1-|x|)$ for $|x|<1$ and zero outside. Its Fourier transform is real and even:At zero the removable limiting value is one, also obtained directly by integrating the triangle. This is the Fourier transform of a triangular function.