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Take transverse length units in which the wave phase modulation has wavenumber one. Set $E_0(z)=e^{i\cos z}$. A Taylor expansion of the free parabolic propagator givesFor a general real wave phase $\varphi$,Multiplying the envelope expansion by its complex conjugate and collecting powers of $x$ gives the phase-to-intensity conversion near a deterministic phase screen,For $\varphi =\cos z$, ${\varphi }^{\prime }=-\sin z$, ${\varphi }^{\prime \prime }=-\cos z$ and ${\varphi }^{\prime \prime \prime }=\sin z$, henceThe first-order brightening occurs where the wave phase curvature is negative. Both correction terms average to zero over a transverse period, in agreement with power conservation. If a dimensional modulation $\cos (qz)$ is used instead, the two terms are $(q^{2}x/k)\cos (qz)$ and $(q^4{x}^2/k^2)\cos (2qz)$.

Treat $\psi$ and its complex conjugate as independent variables in the functional derivative. Integrating the gradient variation by parts, with vanishing boundary terms, givesThe factors follow from ${\partial }_{{\psi }^{\ast }}|\psi {|}^4=2|\psi {|}^2\psi$ and ${\partial }_{{\psi }^{\ast }}|\psi {|}^6=3|\psi {|}^4\psi$. Thus the cubic-quintic Gross–Pitaevskii equation is

Use decaying variations with the same complex bulk field one, so that integration by parts produces no variation boundary term. The functional derivatives of energy and renormalized momentum of a condensate areFor the second identity, variation of the two momentum terms and integration of the ${\partial }_z\delta {\psi }^{\ast }$ term each supply $(2i{)}^{-1}{\psi }_z$, so together they give $-i{\psi }_z$. It follows thatis exactly the stationary Gross–Pitaevskii solitary wave equation. Thus each solitary wave is a constrained energy critical point with multiplier $U$.

Let $s$ parametrize a differentiable family of these waves. Apply the critical-point identity to the variation ${\partial }_s\psi$, keeping the multiplier fixed at its value for that member of the family. Including the complex conjugate variation givesHence the energy–momentum slope of a solitary wave iswherever $dp/ds\ne 0$. At a momentum turning point the parametrized identity $dE/ds=Udp/ds$ remains the correct statement; a globally single-valued energy as a function of momentum is not needed. Differentiating $E-U(s)p$ as a composite function would include an extra $-p{U}^{\prime }$ term, so that composite derivative must not be set to zero.

Define the physical velocity potential by $\psi =\sqrt{n}\exp (im\varphi /\hslash )$, so the superfluid velocity is $\mathbf{v}=\nabla \varphi$. The condensate number current isMultiplying the pumped Gross–Pitaevskii equation by ${\psi }^{\ast }$ and subtracting its complex conjugate gives the density equation. Equivalently, substituting the amplitude and phase and equating imaginary parts yieldsEquating real parts gives the other gain-loss Madelung equations relation,The last term is the quantum potential, retained wherever $n>0$. Taking a gradient gives the equivalent velocity equationin an irrotational region. In equilibrium $n_t=0$ and $\varphi (\mathbf{x},t)=\Phi (\mathbf{x})-\mu t/m$, soThis convention distinguishes the physical velocity potential from the dimensionless quantum phase $m\varphi /\hslash$; using the latter instead would change the divergence prefactor.

The limiting polynomial has a double root at $x=-1$, so two roots require a Puiseux series. For positive $ϵ$, put $\delta =\sqrt{ϵ}$ and $x=-1+a\delta +b{\delta }^2+\cdots$. Substitution and dominant balance giveThus $a=\pm 1$ and $b=-3/2$. The two finite roots have the three-term asymptotic expansionsThis is square-root splitting of a double polynomial root. For negative $ϵ$, the same formulas apply with a chosen complex square root and describe a complex conjugate pair.

The third root is a divergent algebraic root in a singular perturbation. Set $x=X/ϵ$: the equation for $X$ is $X^3+X^2+2ϵX+{ϵ}^2=0$. Its nonzero limiting root is $X=-1$. Substituting $X=-1+Aϵ+B{ϵ}^2+\cdots$ yields $A=2$ and $B=3$, henceAs an independent check, Vieta formulas give the sum of roots as $-1/ϵ$: the two finite-root expansions sum to $-2-3ϵ+O({ϵ}^2)$, giving the same large-root expansion.

Away from the velocity-profile corners, $U^{\prime \prime }=0$ and an unstable normal mode satisfies ${\varphi }^{\prime \prime }-k^2\varphi =0$. On the upper half of the jet writeThe exterior solution is selected by decay at infinity. From the given velocity components, the perturbation pressure isContinuity of normal velocity and pressure therefore gives continuity of $\varphi$ and pressure matching at a piecewise-linear shear interface:At $y=1$ the slope of $U$ jumps from $-1$ to zero. Using ${\varphi }^{\prime }(1^{+})=-k\varphi (1)$ yieldsThe lower interface then follows by parity.

For the even sinuous mode of a planar jet, the derivative can jump at the center: being an even function does not require the one-sided derivative to vanish. The slope jump at $y=0$ is $-2$, whereas $[{\varphi }^{\prime }]=2{\varphi }^{\prime }(0^{+})$. ThusSubstitute $A=-k(1-c)B$ into the upper-interface relation. With $q=e^{-2k}$, division by $e^k$ and multiplication by $2k$ giveCollecting powers of the phase velocity gives the required dispersion relation:These equations can also be treated as a homogeneous two-by-two system in $A,B$, avoiding division by any potentially zero amplitude or phase-velocity factor.

For the odd varicose mode of a planar jet, $\varphi (0)=0$, so $A=0$; the central pressure matching at a piecewise-linear shear interface is automatic. The upper-interface equation becomes $ck{e}^k=\sinh k$, and thereforeThis phase velocity is real for every $k>0$, so the discrete varicose mode of a planar jet is neutral and there is no exponentially unstable odd mode.

To determine the unstable even modes, factor the reduced discriminant of the quadratic:The second factor vanishes at zero and has derivative $2(1-e^{-k}{)}^2>0$ for $k>0$, so it is positive there. The first factor starts at $-8$, is strictly increasing, and tends to infinity. It therefore has one positive zero $k_c$. The even phase velocities areIt follows thatFor $0<k<k_c$, one member of the complex conjugate pair grows at rate $k{c}_i=\sqrt{-D(k)}/4$ and the other decays. At $k=k_c$ the speeds merge and there is no exponential growth; for $k>k_c$ both speeds are real. This is inviscid instability of a triangular jet. The unstable sinuous mode of a planar jet also satisfies Howard's semicircle theorem with $U_{\min }=0$ and $U_{\max }=1$: its speed lies in the upper half-disk centered at $1/2$ with radius $1/2$.

Write $z=x+iy$ and $\alpha =p+iq$. The complex conjugates in the PDF giveSince $(p,q)\ne (0,0)$, the locus is the straight line through the origin perpendicular to the position vector of $\alpha$ in the complex plane. Equivalently,

A branch of the complex logarithm must be specified to obtain a unique locus. First use the principal complex logarithm. Away from the negative real axis, write $z=r{e}^{i\theta }$ with $r>0$ and $-\pi <\theta <\pi$. Then the complex logarithms of $z$ and its complex conjugate are $\log r+i\theta$ and $\log r-i\theta$. Equating real and imaginary parts gives $\log r=\theta$. Consequently the principal-branch locus is the open segment of a logarithmic spiralThe origin is excluded because its complex logarithm does not exist. If the principal value is assigned argument $\pi$ on the negative real axis, a point there equals its own complex conjugate, so its principal complex logarithm would have to equal $i$ times itself. That forces the complex logarithm to be zero and cannot hold there; the endpoints remain excluded.

If the printed equation instead means that some values of the multivalued complex logarithms satisfy it, write those values as $\log r+it$ and $\log r+iu$, with $e^{it}=z/r$ and $e^{iu}=\overline{z}/r$. Equality gives $\log r=-u=t$, hence $u=-t$, which is compatible with conjugation. Every real $t$ is allowed. Under this interpretation the locus is the entire logarithmic spiral $z=e^{(1+i)t}$, $t\in \mathbb{R}$. The PDF does not choose between these branch conventions.

The shifted matrix $N=D-\lambda I$ is still a normal matrix, sinceand $N{N}^{†}$ has the same expansion except that its first term is $D{D}^{†}$, which equals $D^{†}D$. For the given eigenvector, $Ne=0$. Part (b), applied to $N$, therefore gives $N^{†}e=0$. Since $N^{†}=D^{†}-\overline{\lambda }I$,The corresponding eigenvalue of the Hermitian conjugate is the complex conjugate of the original eigenvalue.

For a complex matrix, its Hermitian conjugate is obtained by transposing and taking entrywise complex conjugates:The defining conditions areFor a square matrix, either one of the two inverse identities in the unitary matrix condition implies the other. A Hermitian matrix may have complex entries, but its diagonal entries are real.

For a one-dimensional wavefunction, the probability density and probability current areThe Time-dependent Schrödinger equation with a real potential is $i\hslash {\psi }_t=-{\hslash }^2{\psi }_{xx}/(2m)+V\psi$. Multiply it and its complex conjugate by ${\psi }^{\ast }$ and $\psi$, respectively. Subtracting cancels the real-potential terms and givesThis derives the continuity equation ${\rho }_t+j_x=0$.

For $k>0$, the specified free-region wavefunction combines a right-moving incoming wave of unit amplitude and a left-moving reflected wave with complex reflection amplitude $R$. Applying the free Schrödinger operator to either plane wave gives $E={\hslash }^2{k}^2/(2m)$. Its time factor has modulus one, soThe density includes the stationary interference pattern; $|R{|}^2$, rather than $R$, is the reflection probability.

For the finite square well, let $k=\sqrt{2mE}/\hslash$ outside and $q=\sqrt{2m(E+V_0)}/\hslash$ inside. Write the stationary parts as $e^{ikx}+R{e}^{-ikx}$ to the left, $A{e}^{iqx}+B{e}^{-iqx}$ inside, and $T{e}^{ikx}$ to the right. At both finite potential steps, the wavefunction and its first spatial derivative are continuous. Integrating the stationary Schrödinger equation across a shrinking step proves derivative continuity; a jump of the wavefunction would create an unmatched distributional derivative and is excluded.

Propagate the boundary values at $a$ back through the well. The values at zero areMatching them to $1+R$ and $ik(1-R)$, and adding the two equations after division of the second by $ik$, givesSince the exterior wave numbers agree, the transmission probability is the current ratio $|T{|}^2$, soWhen $V_0=3E$, $q=2k=\sqrt{8mE}/\hslash$, yieldingClassically an incoming particle of positive energy crosses the attractive well and emerges with its original energy, so transmission is one. The quantum result agrees exactly when $qa$ is an integer multiple of $\pi$:These are widths with resonant transmission through a square well; the degenerate width zero also has unit transmission.

A two-component Weyl spinor is a complex spinor transforming in one of the two inequivalent group representations $(1/2,0)$ and $(0,1/2)$ of the Lorentz spinor double cover $\mathrm{SL}(2,\mathbb{C})$. An undotted left-handed field ${\chi }_{\alpha }$ transforms by a matrix $M\in \mathrm{SL}(2,\mathbb{C})$; its complex conjugate ${\bar{\chi }}_{\dot{\alpha }}$ transforms by $M^{\ast }$. Dotted and undotted indices distinguish the two chirality representations. Raising a dotted index uses the invariant epsilon tensor and converts the complex conjugate transformation into the equivalent inverse-adjoint form. Rotations act by the same SU(2) generators on the two chiralities, while boosts have opposite generators, so the two complex representations cannot be identified by a complex-linear Lorentz invariant map.

The antisymmetric invariant tensors ${ϵ}_{\alpha \beta }$ and ${ϵ}_{\dot{\alpha }\dot{\beta }}$ raise and lower spinor indices; $M^TϵM=ϵ$ proves invariance of the contracted bilinear $\chi \psi ={\chi }^{\alpha }{\psi }_{\alpha }$. The components of a fermionic Weyl spinor are odd Grassmann variables, so this bilinear is symmetric under exchange of the two spinors: the antisymmetric epsilon sign cancels the Grassmann sign. In particular, choosing ${ϵ}_{12}=1$ and ${ϵ}^{12}=-1$ gives $\chi \chi =2{\chi }_1{\chi }_2$, generally nonzero. This Weyl spinor bilinear exchange identity permits a Majorana mass term for a single suitable Weyl spinor. With mostly-minus metric signature and ${\bar{\sigma }}^{\mu }=(I,-\mathbf{\sigma })$, the free massless kinetic term is $i{\chi }^{†}{\bar{\sigma }}^{\mu }{\partial }_{\mu }\chi$, and its field equation of motion is $\boxed{{\bar{\sigma }}^{\mu }{\partial }_{\mu }\chi =0}$.

Here L-superfield means a left-chiral superfield: ${\bar{D}}_{\dot{\alpha }}\Phi =0$, and its complex conjugate is an antichiral superfield, $D_{\alpha }{\Phi }^{\ast }=0$. Interpret the two barred derivatives as the contracted scalar ${\bar{D}}^2={\bar{D}}_{\dot{\alpha }}{\bar{D}}^{\dot{\alpha }}$. The key point is chiral projection by squared supercovariant derivatives. Since the two components of $\bar{D}$ anticommute, any product of three barred derivatives repeats a component and is zero. Thusfor every superfield $X$, even when $X$ is antichiral. In particular, ${\bar{D}}^2{\Phi }^{\ast }$ is chiral.

The right side is chiral too. Because $\Phi$ is even, the graded derivation rule gives ${\bar{D}}_{\dot{\alpha }}({\Phi }^2)=({\bar{D}}_{\dot{\alpha }}\Phi )\Phi +\Phi ({\bar{D}}_{\dot{\alpha }}\Phi )=0$. Constant $m$ and $\lambda$ do not contribute derivatives. Consequentlyso applying the chiral constraint to the equation imposes no additional incompatible condition. This is the holomorphic closure of chiral superfields; a generic nonholomorphic right side would not satisfy it.

Both sides have zero Grassmann parity and are Lorentz scalars: the two dotted derivative indices are contracted with the invariant epsilon tensor, and $\Phi$ is a scalar superfield. Moreover, each covariant derivative anticommutes with every supercharge, so its squared contraction commutes with the supercharges. Hence the equation transforms covariantly under supersymmetry. It is the chiral field equation of motion of the Wess–Zumino model with superpotential $\mathcal{W}(\Phi )=m{\Phi }^2/2+\lambda {\Phi }^3/3$, for which ${\mathcal{W}}^{\prime }=m\Phi +\lambda {\Phi }^2$, using the normalization in the PDF. With mass dimensions $[\Phi ]=1$, $[\bar{D}]=1/2$, $[m]=1$ and $[\lambda ]=0$, each side has dimension two. No component field equations of motion are needed for this consistency argument.

The dominant balance for algebraic roots separates one small root of a polynomial from three large ones. For the small root of a polynomial, write $z=\epsilon +a{\epsilon }^3+O({\epsilon }^5)$ in $z=\epsilon (1+z^2{)}^2$. Comparing coefficients gives $a=2$, soThis is a real analytic function of real $\epsilon$ near zero, since the derivative with respect to $z$ of $z-\epsilon (1+z^2{)}^2$ is one at $(0,0)$.

For the large roots of a polynomial, use the real cube root $s=\mathrm{sgn}(\epsilon )|\epsilon {|}^{1/3}$ and set $z=w/s$. The transformed equation isIts three nonzero leading roots of a polynomial are $w=\omega$, with ${\omega }^3=1$. Writing $w=\omega +a{s}^2+O(s^4)$ gives $a=4a+2{\omega }^2$, hence $a=-2{\omega }^2/3$. ThereforeThe choice $\omega =1$ is the second real root of a polynomial; the other two form a complex conjugate pair for either sign of $\epsilon$. The nonzero transformed roots of a polynomial are simple, so their Taylor series in $s^2$ exist locally and exhaust the three large roots of a polynomial. Together with the small one these give all four roots of a polynomial for sufficiently small nonzero $\epsilon$. At $\epsilon =0$ itself the rational equation has only the finite root zero; the other three escape to infinity. Multiplying by $(1+z^2{)}^2$ has not introduced roots at $z=\pm i$, since those points do not solve the multiplied equation.

For $s=2$, part (a) gives $\rho (w)=\frac{3}{2}{w}^2-2w+\frac{1}{2}$ and the right-side polynomial is $w^2$. On the scalar test equation $y^{\prime }=\lambda y$, put $z=h\lambda$. Its characteristic equation isThe linear stability domain contains every $z$ for which both roots have modulus strictly less than one. For real $-1/2\le z<0$, the roots are positive; their larger numerator is less than three while their denominator exceeds three, so both are less than one. At $z=-1/2$ the repeated root is $1/2$, also strictly inside the unit circle. For $z<-1/2$, the roots are complex conjugates and their product isThus in every case both roots lie strictly inside the unit disk. Repeated roots there are allowed, since polynomial factors times their powers still tend to zero. Therefore

The polar form of a complex number and the roots of unity givesince $512=2^9$. Using the complex conjugate to compute the squared complex modulus,The residue $3j-k$ runs through every class modulo $9$, already with $j=0$. The cosine is even, so the nine angles give five distinct values, represented by $r=0,1,2,3,4$. They are distinct because cosine is strictly decreasing on $[0,\pi ]$ and these representative angles lie in that interval. HenceEvery listed value is attained by choosing $z_1=1$ and $z_2=2e^{2\pi ir/9}$.

Away from the negative real axis, the complex conjugate has principal complex argument $-\theta$. ConsequentlyTaking the complex modulus gives $\cosh \theta =1$, hence $\theta =0$. The remaining equation is $e^{i\log r}=i$, soOn the negative real axis, both $z$ and $\overline{z}$ have principal complex argument $\pi$, so the left-hand side has modulus $2e^{-\pi }$, not two. There are no additional solutions there. The printed second term is the principal power of $\overline{z}$, not the complex conjugate of $z^i$.

Let $Mv=\lambda v$ with $v\in {\mathbb{C}}^n$ nonzero. For a real symmetric matrix, $M^{\ast }=M$, where the star denotes the conjugate transpose. Consequently $v^{\ast }Mv$ equals its complex conjugate and is real. The eigenvalue equation givessince $v^{\ast }v>0$. Allowing a complex eigenvector in this argument is essential: reality of an eigenvalue must be proved, rather than assumed in advance.

For a prime $p$, the divisor sum defining the real weight is $\varphi (0)-\varphi (\log p/\log X)$. If $p>X^{1/3}$, the second term vanishes and the first is one. Thus $F(p)=1$, while $F(n)\ge 0$ for every integer because $\varphi$ is real and the sum is squared.

Put $h(x)=e^x\varphi (x)$. With the Fourier transform convention in the question, Fourier inversion givesThe function $h$ is smooth and has compact support. Using integration by parts $k$ times, with zero endpoint terms, provesChoose an integer $k\ge A$ to obtain $|\psi (t)|{\ll }_A|t{|}^{-A}$ for $|t|\ge 1$. In particular every polynomially weighted absolute integral of $\psi$ is finite.

For the double integral, useDifferentiating the second formula gives $\int \psi (t)(1+it)e^{-(1+it)u}dt=-{\varphi }^{\prime }(u)$. absolute convergence from the rapid decay permits Fubini's theorem. The derivative energy constant for a smooth sieve cutoff is thereforeThere is no complex conjugate in this integral: the two factors both become $-{\varphi }^{\prime }(u)$, which is real. The constant is positive; in fact the Cauchy-Schwarz inequality applied to $\varphi (1/3)-\varphi (0)=-1$ gives $c_{\varphi }\ge 3$.

Here is the basic structure of the smooth divisor-square sieve asymptotic, with enough uniform estimates to specify the error. Write $L=\log X$ and $R=X^{1/3}$. Expand the square and count multiples of the least common multiple $[d,e]$ in an arbitrary interval $I$ of length $X$:The sum is effectively restricted to $d,e<R$. Each count is $X/[d,e]+O(1)$, independent of the interval's position, so the total error is $O_{\varphi }(X^{2/3})$.

The Fourier representation of a smooth Selberg weight readsHence, with $z^{\prime }=(1+i{t}^{\prime })/L$,The four local terms correspond to the prime dividing neither divisor, only the first, only the second, or both. This is the Euler product for a smoothed divisor-square correlation. Factor it aswhereThe local factors of $H$ are $1+O(p^{-2+4\eta })$ when $|\mathrm{Re}z|,|\mathrm{Re}{z}^{\prime }|\le \eta$ for small fixed $\eta <1/4$. Thus $H$ is a holomorphic function near $(0,0)$, and evaluating each local factor at zero gives exactly one. Consequently $H(0,0)=1$ and $H(z,z^{\prime })=1+O(|z|+|z^{\prime }|)$ there.

The pole expansion $\zeta (1+w)=w^{-1}+O(1)$ now yieldswhen $|t|,|t^{\prime }|\le L^{1/4}$. Integrating the error against the rapidly decreasing transforms gives $O_{\varphi }(L^{-2})$. The complementary tails are harmless uniformly: absolute values in the original divisor series giveAny prescribed power decay of ${\int }_{|t|>L^{1/4}}|\psi (t)|dt$ is available, so these tails and the tails of the limiting kernel are $O_{\varphi }(L^{-2})$ after choosing sufficiently many applications of integration by parts. The same absolute bound justifies the original exchanges of infinite sums and integrals. Using the evaluated double integral,We conclude, uniformly over all interval locations,This proof does not assume that the interval starts near the origin.

Finally choose this one fixed smooth cutoff. primes above $R$ in $(X_0,X_0+X]$ contribute one each to the nonnegative weight, and at most $R$ primes lie below $R$. Thereforefor all sufficiently large $X$, uniformly in $X_0$. The bounded range $2\le X\le X_1$ follows by increasing the fixed constant and using the trivial interval bound $X+1$. Thus the short-interval prime upper bound from a smooth divisor weight is

The space of test functions is $\mathcal{D}(\mathbb{R})=C_c^{\infty }(\mathbb{R})$: its elements are smooth functions with compact support. For a fixed compact set $K$, put ${\mathcal{D}}_K=\{\phi \in C^{\infty }(\mathbb{R}):\mathrm{supp}\phi \subset K\}$ and use the seminorms $p_m(\phi )={\max }_{0\le j\le m}\Vert {\phi }^{(j)}{\Vert }_{\infty }$. The space of test functions carries the usual test-function inductive limit topology of these spaces. In particular, a sequence converges precisely when its supports eventually lie in one compact set and every derivative converges uniformly. Thus ${\phi }_j\to \phi$ means that a common $K$ contains their supports and $p_m({\phi }_j-\phi )\to 0$ for every $m$.

A distribution is a continuous linear functional on this space of test functions, and ${\mathcal{D}}^{\prime }(\mathbb{R})$ denotes their space. We use complex-linear, bilinear pairings $⟨u,\phi ⟩$. Equivalently, for every compact set $K$ there are $C_K$ and a nonnegative integer $m_K$ such thatThe usual weak convergence of distributions is $u_j\to u$ if $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every test function. This specifies the convergence used below; it does not require convergence in any norm.

The distributional derivative is defined byThe derivative map sends ${\mathcal{D}}_K$ continuously to itself, with $p_m({\phi }^{\prime })\le p_{m+1}(\phi )$. The preceding continuity estimate therefore proves that $u^{\prime }$ is again a distribution. This definition extends the ordinary derivative of a smooth function, by integration by parts.

Choose the translation of a distribution convention ${\tau }_{h}f(x)=f(x-h)$. Its action on a test function isFor fixed $h$, the translated test functions have translated compact support and unchanged derivative sup norms, so this defines a distribution. For the differentiability of distribution translations, apply the Taylor theorem in integral form:All supports lie in one slightly enlarged compact set for $|h|\le 1$, and the same identity for every derivative proves uniform convergence. The continuity of $u$ now givesThe minus sign in the translation parameter is necessary for this convention.

For (a), choose a test function $\eta$ with $\int \eta =1$. If $\int \psi =0$, then $F(x)={\int }_{-\infty }^x\psi (s)ds$ is a test function, since the zero integral makes it vanish beyond both ends of the compact support. If $u_1^{\prime }=0$, then $⟨u_1,\psi ⟩=⟨u_1,F^{\prime }⟩=0$. Decompose $\phi =(\int \phi )\eta +\psi$ to obtainConversely, these constant regular distributions have zero distributional derivative. This also proves the general fact that a distribution with zero derivative is constant.