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Set $y=\tanh z$ and $a=1-k$, so $dy/dz=1-y^2$ and $\hat{w}=(1-y^2{)}^{k/2}{y}^a$. Away from $y=0$, the logarithmic derivative givesFor $c=0$, ${\overline{U}}^{\prime \prime }/\overline{U}=-2(1-y^2)$ and $N^2/{\overline{U}}^2=J(1-y^2)/y^2$. The Taylor–Goldstein equation residual divided by $\hat{w}$ consequently simplifies toThus the neutral mode of the equal-width Hazel model requiresThe qualifier “solution” needs care at the critical level of an internal gravity wave $z=0$. For $0<k<1$, $\hat{w}\sim z^{1-k}$ and ${\hat{w}}^{\prime }\sim (1-k)z^{-k}$: it is a solution separately on either side, not a classical smooth eigenfunction across the critical level of an internal gravity wave. A branch for negative $\tanh z$ is also needed. For example, a limit from $c_i>0$ assigns $(\tanh z{)}^{1-k}=|\tanh z{|}^{1-k}{e}^{-i\pi (1-k)}$ on $z<0$.

The critical-layer regularity of a neutral Hazel mode further gives local finite horizontal kinetic energy only for $0<k<1/2$, since incompressible flow gives $\hat{u}=i{\hat{w}}^{\prime }/k$ and ${\int }_0^{ϵ}{z}^{-2k}dz$ then converges. At $k=1$, $J=0$ and $\hat{w}=\mathrm{sech}z$ is smooth after the removable ${\overline{U}}^{\prime \prime }/\overline{U}$ quotient is continued. At $k=0$, $J=0$ and the formal $\tanh z$ profile does not decay at infinity, so it fails the remote endpoint condition. These qualifications prevent interpreting the whole closed parameter interval as a family of classical decaying modes.

With the sign convention of the displayed operator, the Lax equation is $L_t=[L,A]$. Differentiating $L\phi =k^2\phi$ shows that ${\phi }_t+A\phi$ is another solution at the same spectral parameter. The normalization at $-\infty$ fixessince $A{e}^{-ikx}=4i{k}^3{e}^{-ikx}$ there. At $+\infty$, comparison of the two plane-wave coefficients givesThus the transmission denominator $a$ is time independent. This is the Time evolution of KdV scattering data for the stated normalization.

The logarithmic derivative $S={\phi }_x/\phi +ik$ obeys the Riccati equationIts formal large-$k$ expansion givesThere is an important qualification to the integral identity in the question. For real $k$ with nonzero reflection, $e^{ikx}\phi \sim a+b{e}^{2ikx}$ has no limit at $+\infty$, so the literal improper integral of the exact $S$ need not exist. Continue to $\mathrm{Im}k>0$, where the reflected exponential decays relative to $e^{-ikx}$, and take sufficiently large $k$ away from zeros and with a consistent logarithm. Then the normalization givesThe second equality is an asymptotic expansion, not an asserted convergent series. Time independence of $a$ fixes every coefficient, so each $\int S_{n}dx$ is a conserved quantity of the Korteweg-De Vries equation. The first two nontrivial examples are $-\int udx$ and $\int u^{2}dx$.

For the usual real-valued KdV potential and real formal $k$, write $S=X+iY$ with real coefficient functions. Its imaginary part of the Riccati equation is $Y^{\prime }-2kX+2XY=0$, henceIn the formal expansion, $X$ contains precisely the even-indexed $S_n$ and $Y$ the odd-indexed ones. Rapid decay of $u$ and its derivatives makes every coefficient of the logarithm vanish at both spatial ends. Therefore $\boxed{{\int }_{\mathbb{R}}{S}_{2j}dx=0(j\ge 1)}$. This is a coefficientwise formal identity; it does not claim that the exact real-axis reflected solution has a convergent integral of $X$. The analytically continued scattering identity and the formal expansion are the conventions needed to make the requested argument valid.

Taking the logarithmic derivative of the grey atmosphere temperature law, $4\log T=\log (\tau +2/3)+\text{constant}$, gives

Use a stationary, non-streaming tracer population with spherical symmetry in its full velocity distribution. In particular, the mean velocities vanish, the mixed covariances vanish, and the two tangential components have equal velocity dispersions. WriteHere ${\sigma }_t^2$ is the variance of one tangential component. In the alternative convention using the sum of both tangential variances, the same velocity-anisotropy parameter is $1-({\sigma }_{\theta }^2+{\sigma }_{\varphi }^2)/(2{\sigma }_r^2)$. These definitions agree; they must not be mixed by inserting an extra factor of two.

The time derivative and mean-velocity terms in the supplied Jeans equation vanish, as do its angular mixed-moment terms. Its remaining geometric term is $2\nu ({\sigma }_r^2-{\sigma }_t^2)/r$. Thus the Spherical Jeans equation becomesThe no-streaming assumption excludes a steady radial flow; stationarity of the density alone would not remove its advective terms.

By the shell theorem, the gravitational potential of a spherical mass distribution satisfies $d\Phi /dr=GM(<r)/r^2$. Expanding the derivative above, dividing by $\nu$, and using logarithmic derivatives givesThe tracer density $\nu$ need not be the density of the gravitating matter. The circular speed is $V_c^2=rd\Phi /dr=GM(<r)/r$, whether or not the tracers themselves follow circular orbits.

Now let $\beta$ be constant, $\nu ={\nu }_0(r/r_0{)}^{-\gamma }$, and $V_c^2=V_0^2(r/r_0{)}^{2\alpha }$. With $y={\sigma }_r^2$, the Spherical Jeans equation readsFor a power-law ansatz $y=K{V}_c^2$, $y^{\prime }=2\alpha y/r$, so $(\gamma -2\beta -2\alpha )y=V_c^2$. Therefore the scale-free spherical Jeans solution isThe positivity condition is necessary for this particular solution to be a physical velocity dispersion.

There is a boundary condition implicit in selecting this solution. The integrating factor $r^{2\beta -\gamma }$ gives, when $D=\gamma -2\beta -2\alpha \ne 0$,The homogeneous term represents an additional radial-pressure boundary contribution. It is removed, for example, by requiring $r^{2\beta }\nu {\sigma }_r^2\to 0$ at infinity. Under that condition,and convergence requires $D>0$, giving the boxed result. If $D=0$, the solution instead has the logarithmic formso the quoted constant proportionality does not apply. A finite outer boundary can retain a homogeneous term even when $D\ne 0$.

To derive the heliocentric projection of a spherical velocity dispersion, put the Sun at vector ${\mathbf{R}}_{\odot }$ of length $R_{\odot }$, and a tracer at $\mathbf{r}=r\hat{\mathbf{r}}$. Its heliocentric line of sight isIf $\chi$ is the angle between $\hat{\mathbf{\ell }}$ and the Galactocentric radial direction, thenAfter correction for the Sun's motion, the intrinsic covariance matrix has eigenvalues ${\sigma }_r^2,{\sigma }_t^2,{\sigma }_t^2$ in the local spherical basis. Projection onto the line of sight therefore givesAt fixed $r$, ${\sigma }_r$ and $\beta$ are constant across the spherical shell. Averaging its squared line of sight velocity, with zero mean, givesFor a uniformly sampled Galactocentric shell, the solid angle measure makes $u$ uniform on $[-1,1]$, giving the geometric correction for heliocentric velocity dispersionTo evaluate it, set $a=r^2+R_{\odot }^2$ and $b=2r{R}_{\odot }$, and divide the integrand:The odd term integrates to zero, while ${\int }_{-1}^{1}du/(a-bu)=b^{-1}\ln [(a+b)/(a-b)]$. For $r>0$, $R_{\odot }>0$, and $r\ne R_{\odot }$, this givesThe removable limit at $r=R_{\odot }$ is $A=1/2$. As $r/R_{\odot }\to \infty$, $A\simeq 2R_{\odot }^2/(3r^2)$, while as $r/R_{\odot }\to 0$, $A\to 2/3$. If $R_{\odot }=0$, $A=0$ and the observed velocity dispersion is exactly radial. For isotropic velocities, $\beta =0$, the correction vanishes at every radius.

The averaging measure matters. If directions are instead uniform in heliocentric solid angle at a fixed Galactocentric radius $r>R_{\odot }$, let $\vartheta$ be the angle between the line of sight and ${\mathbf{R}}_{\odot }$. Geometry gives ${\sin }^2\chi =(R_{\odot }/r{)}^2{\sin }^2\vartheta$, so that particular weighting gives $A=2R_{\odot }^2/(3r^2)$ exactly. The logarithmic expression above uses uniform tracer weighting on a Galactocentric shell. For an observational sample, use its actual angular selection weights in $A=⟨{\sin }^2\chi ⟩$; the boxed dispersion relation remains valid.

Shrink the constant $c$ from part b if necessary. Put $L=\log |t|$. Zeros in the disk appearing in the supplied partial-fraction formula have $\log |\gamma |=L+O(1/|t|)$, so part b ensuresfor every such zero.

If $\sigma \ge 1+c/(2L)$, absolute convergence of the logarithmic derivative giveswith the region $\sigma \ge 2$ even easier.

It remains to take $1-c/(2L)<\sigma <1+c/(2L)$. SetFor every local zero, both $\Re (s-\rho )$ and $\Re (s_0-\rho )$ are positive and comparable, while $|s-s_0|\ll 1/L$. The partial-fraction formula at $s_0$, together with the preceding Euler-product bound, givesSince $\Re (s_0-\rho )\gg 1/L$, it follows thatSubtracting the partial-fraction formulas at $s$ and $s_0$ now yieldsTherefore the logarithmic derivative inside the zeta zero-free region satisfiesthroughout the required half-width region.

Assume first the Riemann hypothesis. Replace any $x\ge 2$ by $y=\lfloor x\rfloor +1/2$; then $\psi (x)=\psi (y)$ and $⟨y⟩\ge 1/2$. Take $T=y$ in part a. Every nontrivial zero has real part $1/2$, and the Riemann–von Mangoldt formula impliesConsequentlywhile both truncation errors in part a are $O((\log y{)}^2)$. Thuswhich implies the stated $O_{ϵ}(x^{1/2+ϵ})$ estimate.

Conversely, suppose that estimate holds for every $ϵ>0$. For $\Re s>1$, partial summation givesGiven any $s$ with $\Re s>1/2$, choose $ϵ<\Re s-1/2$. The error hypothesis makes the last integral locally uniformly convergent there, so it supplies a holomorphic continuation ofto the half-plane $\Re s>1/2$. A zero of $\zeta$ in that half-plane would create a pole of its logarithmic derivative, so none exists. The functional equation of the Riemann zeta function reflects every nontrivial zero with real part below $1/2$ to one above $1/2$. All nontrivial zeros must therefore lie on the critical line, proving the Riemann hypothesis equivalence for the second Chebyshev function.

The Mellin transform of $F$ isBecause the support is a compact subset of $(0,\infty )$, the integral defines an entire function of $s$. The Mellin inversion formula says that, for every real $\sigma$ and every $y>0$,

Apply this with $y=n/X$ and initially $\sigma =2$. Absolute convergence of the Dirichlet series for the logarithmic derivative permits interchange of sum and integral, givingTruncate at height $T=(\log X{)}^A$, where $A$ is a sufficiently large fixed constant. The assumed bound on $\tilde{F}$ makes the discarded tails smaller than the required error. The classical zero-free region of the Riemann zeta function and the standard bound ${\zeta }^{\prime }(s)/\zeta (s)\ll (\log (|t|+3){)}^{O(1)}$ there allow the truncated contour to move toThe only singularity crossed is the simple pole of $-{\zeta }^{\prime }/\zeta$ at $s=1$, whose residue is $1$. Its contribution isOn the new contour, $|X^s|=X^{1-c/\log \log X}$; the zeta bounds, contour length, and exponential decay of $\tilde{F}$ absorb into a slight decrease of $c$. ThereforeThis is the smoothed prime number theorem from a zero-free region.

The truncated Perron formula says that if $F(s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$ converges absolutely for $\Re s>{\sigma }_a$, then for $c>{\sigma }_a$, $T\ge 2$, and $x$ not an integer,Take $a_n=\Lambda (n)$ and $c=1+1/\log x$. The logarithmic derivative identity gives $F(s)=-{\zeta }^{\prime }(s)/\zeta (s)$. Since $x-1/2$ is an integer, $|x-n|\ge 1/2$ for every integer $n$. In the range $x/2<n<2x$,and hence the contribution there isThe ranges $n\le x/2$ and $n\ge 2x$ are bounded by the same quantity using absolute convergence and $-{\zeta }^{\prime }(c)/\zeta (c)\ll \log x$. Therefore

Put $t=s-1$. The Laurent series of the logarithmic derivative at the simple pole of $\zeta$ has the formwhere in fact $c_0=\gamma$. HenceThe coefficient of $t^{-1}$ in their product, and therefore the residue, iswith the constant $A=-1-2c_0$; equivalently, $A=-1-2\gamma$.

A periodic solution is a solution of a differential equation which is a periodic function: for some $T>0$, $x(t+T)=x(t)$ at every time. This demands repetition of the state, rather than merely repetition of the coefficients in the differential equation. A nonconstant periodic solution of an autonomous differential equation traces a periodic orbit; constant equilibrium points are also periodic functions, so they must be excluded when asserting absence of periodic orbits.

For an explicit example, consider the initial value problemIts coefficient has period $T=2\pi$. Taking a logarithmic derivative and then applying integration givesDirect differentiation verifies both the differential equation and the initial condition. After one coefficient period,Thus the repeating coefficient generally produces a repeating shape with a changed amplitude. With $b=1$, the three cases $a=-0.2,0,+0.2$ have one-period factors approximately $0.2846,1,3.5136$, respectively. Only $a=0$ gives a periodic solution with this initial condition. For $a<0$ the positive solution of a differential equation tends to zero, and for $a>0$ it tends to infinity; neither can have any positive period. The zero solution of a differential equation, obtained with a different initial condition, is periodic for every $a$.

The exponential function separates two roles:For $b=1$, even the decaying case $a=-0.2$ initially rises, because $y^{\prime }(0)=0.8>0$. The growing case $a=0.2$ has intervals of decline, because $a+\cos t$ becomes negative. This is the distinction between the instantaneous logarithmic derivative $y^{\prime }/y$ and the Floquet growth rate measured over many periods.

More generally, for a real continuous periodic function $q$ of period $T$, the scalar linear differential equation $y^{\prime }=q(t)y$ hasA nonzero solution of a differential equation is $T$-periodic exactly when $m=1$, or equivalently when the coefficient has zero integral over a period. This scalar calculation is the simplest example of a Floquet multiplier.^[ref]

Differentiate the genus-one Hadamard factorization. Its complex difference terms converge as $O(|\rho {|}^{-2})$, and the individual real sums converge by the absolute convergence of the real xi logarithmic derivative. The remaining constant is $\Re B+\sum \Re (1/\rho )$. On the critical line the xi logarithmic derivative has zero real part, while the kernel terms cancel in pairs reflected across that line. Hence this constant vanishes and gives the displayed formula away from zeros.

For $c=1+1/\log x$, $1\le y\le x$ and $2\le T\le x$, apply the truncated Perron formula with Von Mangoldt function coefficients at both endpoints on the same vertical line. The difference kernel is $((x+y{)}^s-x^s)/s={\int }_x^{x+y}{u}^{s-1}du$, of modulus $O(y)$. The usual near-integer error bound contributes $O(x{\log }^{2}x/T)$. The logarithmic derivative on this line is finite because of the Euler product.

Evaluate the real logarithmic derivative of Riemann xi at real part two, where every summand is positive and comparable to this kernel. The gamma logarithmic derivative and the bounded zeta logarithmic derivative give the logarithmic bound. It implies the local unit-interval count, and conversely such local counts imply this smoothed bound by summing the decaying tails.

The boundary can recover more than a count. Under the hypotheses of the argument principle, let $h$ be a holomorphic function on a neighborhood of the contour and its interior. If the enclosed zeros of a function are $a_j$ with zero multiplicities $m_j$ and the enclosed poles are $b_k$ of orders $n_k$, thenMultiplying the local logarithmic derivative by $h$ changes its residue from $m_j$ to $m_{j}h(a_j)$, or from $-n_k$ to $-n_{k}h(b_k)$. The residue theorem then gives the identity. Choosing $h=1$ recovers the argument principle; choosing $h(z)=z$ gives the sum of the enclosed zero locations minus the sum of the enclosed pole locations, both counted with zero multiplicities or pole orders.^[ref]

For the rational function in argument principle for a rational function, the $h(z)=z$ integral is $2(1/4)-(-1/2)=1$. Adding the known pole location $-1/2$ back gives a zero-location sum of $1/2$. Dividing by the zero count two gives the arithmetic mean of the enclosed zeros, with multiplicity, as $1/4$. Choosing $h(z)=z^k$ similarly gives the difference between the power-sum symmetric polynomials evaluated on the enclosed zero and pole locations, repeating each location according to its multiplicity or order. These identities let boundary information constrain both the number and aggregate locations of enclosed zeros of a function.

The classical zero-free region asserts that for some $c>0$,Together with bounds for the logarithmic derivative, it permits contour arguments with exponentially small errors in $\sqrt{\log x}$.