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In the razor-thin disk approximation, the three-dimensional Poisson equation isFor a horizontal Fourier mode with wavevector $\mathbf{k}$ and $k=|\mathbf{k}|>0$, the Fourier transform of this equation isThe solution that decays away from the disk isConsequently the required razor-thin disk Poisson kernel in the midplane isThe spatially uniform $k=0$ background is excluded from this local perturbation formula.

Take the Fourier transform in $x$, with convention $\tilde{v}(k_x)=\int v(x)e^{-i{k}_{x}x}dx$. The differentiation rules $\widehat{v^{\prime \prime }}=-k_x^2\tilde{v}$ and $\widehat{x^{2}v}=-{\tilde{v}}^{\prime \prime }$ turn the forced equation intoFor the unforced equation, $d{k}_x/dt=S{k}_y$ impliesIt is therefore exactly the shearing-wave oscillator found in part c, now parametrized by radial wavenumber rather than time.

In free space $Q=0$, and the fourth moment obeysFor the two-dimensional Fourier transformthe Fourier transform of a derivative turns this equation into the ordinary differential equationThusUsing the screen-exit value from part (a), the inverse transform givesthrough first order in $\Delta x$. Equivalently, the free-space fourth-moment propagator has kernel

The Fourier transform sends each spatial derivative to $i{q}_i$, so at Gaussian levelFor $\kappa <0<\gamma$, minimizing over $s=q^2\ge 0$ givesThe minimum kernel is $G(q_0)=a-{\kappa }^2/(4\gamma )$. Gaussian fluctuations first diverge when this vanishes, so the nonzero-wavevector soft-mode sphere becomes unstable at

For $s>\frac{1}{2}$, the Sobolev trace theorem states that the restriction map initially defined on smooth functions extends uniquely to a bounded linear operator

Take first $u$ in the Schwartz space. Up to the harmless constant determined by the Fourier transform convention,The Cauchy-Schwarz inequality, with weights $1+|{\xi }^{\prime }{|}^2+{\xi }_n^2$, givesAfter the substitution ${\xi }_n=(1+|{\xi }^{\prime }{|}^2{)}^{1/2}t$, the second factor iswhere finiteness is exactly the condition $s>1/2$. Multiplying by $(1+|{\xi }^{\prime }{|}^2{)}^{s-1/2}$ and integrating in ${\xi }^{\prime }$ yieldsDensity of the Schwartz space in the Sobolev space completes the unique bounded extension.

For $f\in L^1({\mathbb{R}}^n)$, use the angular-frequency conventionThis is the Fourier transform on $L^1({\mathbb{R}}^n)$.

Setting $ϵ=0$ makes the inhibitor equation quasistatic:Taking the spatial Fourier transform and using the Fourier transform of a derivative givesTherefore the fast-inhibitor elimination in a reaction-diffusion system is

Taking the Fourier transform of the distributional equation givesbecause $2022$ is even. The multiplier therefore gives the regularity gain for one plus an even power of the Laplacian

The Sobolev embedding theorem in its derivative form says thatTo make all derivatives through order $4044$ continuous, and simultaneously make $f\in H^r$ continuous, it is enough and in general necessary to haveUnder this condition the distributional identity is a pointwise identity, so $u$ is a classical solution. ThusAt the borderline $n=2r$, the standard Sobolev embedding does not in general give continuity.

Let the Bravais lattice have primitive vectors $a_1,a_2,a_3$, and putThe reciprocal lattice has the basisThe scalar triple product and cross product identities giveConsequently every $q={\sum }_j{n}_j{b}_j$ satisfies $q\cdot l\in 2\pi \mathbb{Z}$ for every lattice point $l={\sum }_i{m}_i{a}_i$, and conversely these three conditions force the coefficients of $q$ in the reciprocal basis to be integers. This proves that the displayed vectors generate ${\Lambda }^{\ast }$.

In the Born approximation, suppose that the crystal potential is the sum of translates of one atomic potential $V_0$ over the finite set $S$ of lattice sites. Its Fourier transform at the momentum transfer $Q=k-k^{\prime }$ factors asThus the single-atom scattering amplitude is multiplied by the crystal lattice structure factor $\Delta (Q)$. Writingseparates the sum into three finite geometric series. For $l_i=-L_i/2,\dots ,L_i/2$ this givesWhen the $L_i$ are large, a factor has a sharp maximum when ${\alpha }_i=2\pi n_i$. All three factors are therefore simultaneously large precisely when $Q\in {\Lambda }^{\ast }$. These are the reciprocal-lattice peaks of a finite crystal.

For the stated body-centered cubic lattice basis,the reciprocal-basis formula givesEquivalently,The shortest nonzero reciprocal vectors have squared Euclidean norm $2(2\pi /a{)}^2$, so

In elastic scattering, $|k|=|k^{\prime }|=k$. If $q=k-k^{\prime }\in {\Lambda }^{\ast }$ and $\theta$ is the angle between $k$ and $k^{\prime }$, Euclidean geometry givesThe first possible diffraction peak therefore occurs atas $ka\to \infty$, by the small-angle approximation. This is the elastic Bragg scattering condition for the shortest reciprocal-lattice vector.

The Fourier modes evolve according to the dispersion relation found in part (a), soFor $V>U$, the stationary-point equation isIt has the two real solutionsDefineThe one-dimensional stationary-phase formula gives one contribution from each stationary point:Because the initial field is real, its Fourier transform has the conjugate symmetry $A(-k)=A(k{)}^{\ast }$. Hence the same result can be written

A momentum-shell renormalization group step has three parts. First split the Fourier transform of the field into slow modes ${\varphi }_{<}$ with $|k|<\Lambda /\zeta$ and fast modes ${\varphi }_{>}$ with $\Lambda /\zeta <|k|<\Lambda$, then perform the functional integral over ${\varphi }_{>}$. Second rescale momenta by $k^{\prime }=\zeta k$, equivalently coordinates by $x^{\prime }=x/\zeta$, to restore the cutoff from $\Lambda /\zeta$ to $\Lambda$. Third rescale the field so that the coefficient of $(\nabla \varphi {)}^2/2$ again has its chosen normalization. The effective free energy contains every operator allowed by the symmetries, with transformed coefficients. Repeating the step composes these coefficient maps and produces a renormalization-group flow.

The inverse Fourier transform is linear:Splitting the integration domain into $p^2<b^2{\Lambda }^2$ and $b^2{\Lambda }^2<p^2<{\Lambda }^2$, then inserting the definitions of $\tilde{\varphi }$ and $\tilde{\chi }$, gives

The Schwartz space on the real line isThese seminorms define its Fréchet space topology. The space of tempered distributions is its continuous dual space,

With the angular-frequency convention, the Fourier transform of a Schwartz function isIt maps $\mathcal{S}$ continuously to itself. The transform of $u\in {\mathcal{S}}^{\prime }$ is defined through the dual pairing:up to the fixed reflection and $2\pi$ factor if the inverse-transform convention is used for the test function.

Remove the two reciprocal tails by settingThe assumed $O(x^{-2})$ remainder at each end and local integrability on bounded intervals imply $r\in L^1(\mathbb{R})$. Its Fourier transform $\hat{r}$ is therefore continuous at zero. Since $\widehat{u(-\cdot )}(\lambda )=\hat{u}(-\lambda )$,Parts (iv) and (v) show that both requested one-sided limits exist. If $L_{+}$ denotes the limit from positive frequencies and $L_{-}$ the limit from negative frequencies, then the common value $\hat{r}(0)$ cancels andThis is the universal jump caused by reciprocal tails.

For a test function $f\in \mathcal{D}(\mathbb{R})$, define the proposed integral by reversing the order of integration:The Fourier transform of a test function is a Schwartz function, while $\sqrt{{\theta }^2+1}\asymp 1+|\theta |$. The final integral is consequently absolutely convergent. Repeated integration by parts in $x$ bounds it by finitely many seminorms of the test function, so it defines a continuous linear functional on $\mathcal{D}(\mathbb{R})$.

Equivalently, put $\lambda =\sqrt{{\theta }^2+1}$ on the two half-lines. ThenThe density has only an integrable inverse-square-root singularity at one and is bounded at infinity, so it is a regular tempered distribution. Its inverse Fourier transform is exactly the proposed oscillatory integral. Thus

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, ${\epsilon }_{-k}=-{\epsilon }_k$.