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Set $\epsilon =1-\sigma \in [0,1)$ and $L=\log (t+2)$. By the Hardy-Littlewood approximation to the Riemann zeta function at $x=t$, it is enough to bound ${\sum }_{n\le t}{n}^{-\sigma -it}$: the integral term has size $O(t^{-\sigma })$ because $|s-1|\ge t$.

On a dyadic interval $(N,N^{\prime }]\subset (N,2N]$ with $N^{\prime }\le t$, the assumed estimate holds for every initial subinterval. Abel summation with $n^{-\sigma }$ therefore givesIndeed the weighted endpoint and integral of the term proportional to the subinterval length are $O(N^{1-\sigma })$, and those of the constant $N^{4/5}$ term are $O(N^{4/5-\sigma })$. The constants can be uniform in $0<\sigma \le 1$.

For the first term, write $u=j\log 2$ and complete the square:The sum of a shifted Gaussian function on a fixed-spaced lattice is $O(\sqrt{L})$, uniformly in the shift. Thus these dyadic contributions are $O(\sqrt{L}{e}^{{\epsilon }^{2}L/(4c)})$. This is the Gaussian dyadic summation bound.

For the second term, if $\epsilon \le 1/10$ its dyadic sum is bounded. If $\epsilon >1/10$, a crude bound is $O(L\exp (\max (\epsilon -1/5,0)L))$. The positive exponent obeys $\epsilon -1/5\le (5/4){\epsilon }^2$, and $L$ can be absorbed into $\sqrt{L}{e}^{C{\epsilon }^{2}L}$ uniformly on $\epsilon >1/10$ by increasing the fixed constant $C$. The finitely many initial terms cause no problem. We conclude, with one fixed sufficiently large $C$,In particular the endpoint $\sigma =1$ gives $O(\sqrt{\log t})$ under the assumed exponential-sum hypothesis. This conditional conclusion uses that hypothesis, not an unconditional improvement of the stated Richert bound for the Riemann zeta function.

Choose $f_0(x,v)=e^{-(x^2+v^2)}$ and $h(t,x,v)=x^2+v^2$. The initial Gaussian function is smooth and belongs to every finite Lp space, and is also bounded. The source is smooth and nonzero. The accumulated source along the backward characteristic curve isIts quadratic-form eigenvalues are $(e^{2t}-1)/2$ and $(1-e^{-2t})/2$. Both are positive for $t>0$, soConsequently $\Vert f_t{\Vert }_p=\infty$ for every $t>0$ and every finite $p\ge 1$; it is unbounded, so its $L^{\infty }$ norm is infinite as well. This spatially nonintegrable forcing in transport example avoids any ambiguity about whether the last endpoint is included in “all $p$”.

Use the genuine planar Givens rotationIn the second component the cosine multiplies $v_j$: the repeated $v_i$ in the printed formula is an error. With that printed expression, at $\theta =0$ the pair becomes $(v_i,v_i)$, which does not preserve length or measure. The rotation-based claims require the corrected expression. Also take $N\ge 2$, since the normalization by $\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ is undefined for $N=1$.

Let $C_N=\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ and $U_{ij,\theta }F=F\circ R_{ij,\theta }$. The change of variables formula and determinant one give $\Vert U_{ij,\theta }F{\Vert }_2=\Vert F{\Vert }_2$. Thus each is a unitary operator, with adjoint $U_{ij,-\theta }$. The Kac collision operator is the averageThe Minkowski integral inequality gives $\Vert QF{\Vert }_2\le \Vert F{\Vert }_2$, so $Q$ is bounded. For the Hilbert space inner product, integration and the angular change $\theta \mapsto -\theta$ giveHence $Q=Q^{\ast }$ and $\Vert Q\Vert \le 1$. In fact a nonzero radial Gaussian function is fixed by every rotation, showing $\Vert Q\Vert =1$. Angular averages can be understood as strong Bochner integrals; continuity of rotations in $L^2$ follows first for smooth compactly supported functions, then by density.

Let $h=\log \psi$, with $\psi >0$, and set $f=2\alpha h_{\theta }$. Direct differentiation givesThe heat equation ${\psi }_Z=\alpha {\psi }_{\theta \theta }$ implies $h_Z=\alpha (h_{\theta \theta }+h_{\theta }^2)$, so the bracket vanishes. Conversely its being independent of $\theta$ can be absorbed into a $Z$-dependent multiplicative normalization of $\psi$, leaving $f$ unchanged. This proves the Cole-Hopf transformation with the positive sign appropriate to the negative-flux Burgers convention. Although the algebra works for nonzero $\alpha$, the given forward Gaussian function diffusion kernel and a physical vanishing-viscosity limit require $\alpha >0$.

Take $U,L>0$ for the stated Burgers N-wave. Integrating ${\partial }_{\theta }\log \psi (0,\theta )=f_0(\theta )/(2\alpha )$ and normalizing the exterior value to one givesThis function is continuous at $\pm L$; its logarithmic derivative has the specified jumps. Convolution with the heat kernel is positive and solves the heat equation for $Z>0$. Splitting the integral into the exterior baseline and the interior correction yieldsPut $a=1+UZ$. Completing the square givesChanging the interior integration variable to $\eta =a\phi$ changes its limits to $\pm La$ and the Gaussian function width to $Za$. The Jacobian and normalization leave the factor $a^{-1/2}$. Thus the Cole-Hopf solution for a Burgers N-wave isHere $I_{\alpha }$ is the normalized Gaussian function mass of its indicated interval. Its explicit error function representation isFor fixed $w>0$, the Gaussian function concentrates at $\phi =\theta$ as $\alpha \downarrow 0$. Consequentlyaway from the endpoints; at $\theta =\pm L$ the limit is $1/2$. The transition layer has width $O(\sqrt{\alpha w})$. This is an approximate identity argument, not a uniform step approximation across the endpoints.

For fixed $Z>0$ and $|\theta |\ll L$, both interval masses tend to one, and the exponentially large positive $\hat{\psi }$ dominates the exterior correction. Its logarithmic derivative therefore givesFor $|\theta |\gg La$, both interval masses are exponentially small and the weighted interior integral is also negligible, while the exterior contribution tends to one. Therefore $\boxed{f(Z,\theta )\simeq 0}$ in the specified far exterior.

An exponentially weighted Gaussian function tail should not be discarded solely because its unweighted interval mass tends to zero. In fact, comparing the order-one exterior term with $\hat{\psi }$ gives the sharper inviscid Burgers N-wave fronts $|\theta |=L\sqrt{1+UZ}$, not $La$. Away from these fronts, the vanishing-viscosity limit isInside $|\theta |<La$, this follows from the sign of $L^2-{\theta }^2/a$ in the exponential. Outside $La$, the constrained Gaussian function maximum lies at an interval endpoint and has negative exponent. The right shock wave's speed is $UL/(2\sqrt{a})$, equal both to the derivative of $L\sqrt{a}$ and to minus half the sum of its two limiting states; the left shock wave is its reflection. This checks consistency with the Rankine-Hugoniot condition and shows that the requested near-center and far-exterior approximations are compatible with the full entropy limit.

For a full-rank Euclidean lattice $\Lambda \subset {\mathbb{R}}^n$, its dual lattice isIf $\Lambda =B{\mathbb{Z}}^n$, then ${\Lambda }^{\prime }=B^{-\mathsf{T}}{\mathbb{Z}}^n$, $m(\Lambda )=|\det B|$ and $m({\Lambda }^{\prime })=m(\Lambda {)}^{-1}$. The characters of a real torus identify the additive dual lattice with the multiplicative character group byThe map is well defined precisely because $u\in {\Lambda }^{\prime }$, and is injective. For surjectivity, a continuous group homomorphism from the compact torus into ${\mathbb{C}}^{\times }$ has compact image. Its modulus has logarithm a homomorphism into $(\mathbb{R},+)$ with compact image, hence is zero, so the image lies in the unit circle. Pull the character back to ${\mathbb{R}}^n$. Its continuous real lift under $t\mapsto e^{2\pi it}$, normalized to zero at the origin, is additive: its additive defect is an integer-valued continuous function and vanishes at the origin. A continuous additive function is $⟨u,x⟩$ for a unique $u$. Triviality on $\Lambda$ says $u\in {\Lambda }^{\prime }$, proving surjectivity.

Use the Fourier transform convention $\hat{f}(u)={\int }_{{\mathbb{R}}^n}f(x)e^{-2\pi i⟨u,x⟩}dx$, and take $f$ to be a Schwartz function. The periodization of a Schwartz function $P_f(x)={\sum }_{\lambda \in \Lambda }f(x+\lambda )$ is smooth and $\Lambda$-periodic. On a fundamental cell $F$, the coefficient of the torus character ${\chi }_u$ isThe equality follows by translating each cell and using $e^{2\pi i⟨u,\lambda ⟩}=1$. The rapidly convergent Fourier series can be evaluated at zero, yielding the Poisson summation formula for a Euclidean latticeThis argument keeps track of the covolume factor rather than tacitly assuming a unit-volume lattice.

For $\tau$ in the complex upper half-plane, let $g_{\tau }(x)=e^{\pi i\tau |x{|}^2}$. Scaling the self-dual real Gaussian function gives its Fourier transform at $\tau =it$, $t>0$, and holomorphic continuation in $\tau$ gives the complex Gaussian Fourier transformThe branch is $\exp [-(n/2)\mathrm{Log}(-i\tau )]$ with the logarithm on the right half-plane; it is positive for $\tau =it$. Both the integrals and the lattice sums are locally normally convergent on the complex upper half-plane. Applying the Poisson summation formula proves the lattice theta functional equationNo integrality or self-duality hypothesis on the lattice is needed.

Put $a=n/2$, $m=m(\Lambda )$ and ${\theta }_{\Lambda }(t)={\Theta }_{\Lambda }(it)$. The Epstein zeta function converges absolutely for $\mathrm{Re}s>a$, and its Mellin transform representation isAt infinity, ${\theta }_{\Lambda }(t)-1$ decays exponentially. Define the entire functionOn $(0,1)$, substitute ${\theta }_{\Lambda }(t)=m^{-1}{t}^{-a}{\theta }_{{\Lambda }^{\prime }}(1/t)$ and then $u=1/t$. Isolate the two elementary terms before integrating; this gives the pole-subtracted theta integral for an Epstein zeta functionThe formula initially holds for $\mathrm{Re}s>a$ and continues the completed Epstein zeta function meromorphically to all $\mathbb{C}$. Apply the same formula to the dual lattice at $a-s$, using $m({\Lambda }^{\prime })=m^{-1}$ and ${\Lambda }^{\prime \prime }=\Lambda$. The entire terms and the two rational terms match, provingFinally $E_{\Lambda }(s)={\pi }^s{\mathcal{E}}_{\Lambda }(s)/\Gamma (s)$ also continues meromorphically. Its only pole is a simple one at $s=n/2$, with residue ${\pi }^{n/2}/[m(\Lambda )\Gamma (n/2)]$; the pole of the completed function at zero is cancelled by $1/\Gamma (s)$, and $E_{\Lambda }(0)=-1$. The residues and this cancellation make explicit why subtracting the constant theta term was necessary.

Use the Bromwich inversion formula in the real variable $\omega$, with $p=\gamma +i\omega$:When the relevant Laplace transforms and spatial integral transforms are explicit, every sample of this integrand can be evaluated directly, without a time-stepping approximation to the partial differential equation. Split the initial condition integral at $y=x$: the two pieces of $R_q$ are linear combinations of $e^{\pm qy}$, so truncated spatial exponential transforms suffice. For real data the negative-frequency half is the complex conjugate of the positive-frequency half.

A practical numerical integration is to truncate to $|\omega |\le \Omega$, apply an adaptive quadrature rule, and independently increase $\Omega$ and refine the quadrature. Choose $\gamma >0$ large enough to remain to the right of all singularities, but avoid an unnecessarily large $e^{\gamma t}$. Evaluate hyperbolic function ratios in scaled form; for example,This prevents overflow when $\mathrm{Re}qL$ is large. The boundary contributions at an interior point are damped by $e^{-x\mathrm{Re}q}$ or $e^{-(L-x)\mathrm{Re}q}$, with $\mathrm{Re}q\sim \sqrt{|\omega |/2}$. Near an endpoint more frequencies are needed; at the endpoint itself use the prescribed Dirichlet boundary condition. The initial condition resolvent has a generally only $O(1/\omega )$ tail. Treat its oscillatory tail accurately, subtract a known transform with the same leading term, or evaluate that contribution separately with the heat kernel.

An equally useful check is the eigenfunction expansion. DefineThe causal modal formula isIt follows either from the heat kernel formula or directly from integration by parts against a sine eigenfunction. Exponential-transform formulas evaluate the time integral explicitly when available. The initial condition contribution has a Gaussian function cutoff in $n$ for each positive time. Nonzero endpoint forcing produces a more slowly convergent sine series tail. A boundary lifting $\ell (x,t)=(1-x/L)g_0(t)+(x/L)e^{aL}h(t)$ gives $w=\ell +z$, where $z$ has zero Dirichlet boundary conditions and forcing $-{\ell }_t-a^2\ell$; expanding $z$ gives better convergence and imposes the endpoint values exactly. At very small times the method of images is often more efficient than retaining many modes.

Check convergence by increasing both the frequency cutoff and quadrature resolution, and compare with a separately truncated causal modal or image-kernel evaluation. Causality in finite-time Laplace contour inversion means a leftward contour deformation must respect both resolvent operator poles and growth of the transformed data. In particular, the transform of data extended by zero after $T$ can grow like $e^{-pT}$ in the left half-plane: for $t<T$ one cannot close that contour indiscriminately and discard its large arc. The modal forcing integral with upper limit $t$ avoids this causality error.

First ${\int }_{\mathbb{R}}{e}^{-x^2/2}dx=\sqrt{2\pi }$: square the positive integral and use polar coordinates to obtain $2\pi {\int }_0^{\infty }r{e}^{-r^2/2}dr=2\pi$. Scaling therefore normalizes the one-dimensional Gaussian function $g_t$.

Let $I(\xi )={\int }_{\mathbb{R}}(2\pi t{)}^{-1/2}{e}^{-x^2/(2t)}{e}^{-i\xi x}dx$. Differentiation under the integral is justified by the integrable majorant $|x|g_t(x)$. Since $g_t^{\prime }(x)=-x{g}_t(x)/t$, integration by parts, with zero endpoint terms, gives $I^{\prime }(\xi )=-t\xi I(\xi )$. With $I(0)=1$, this yields $I(\xi )=e^{-t{\xi }^2/2}$. Multiplication of the one-dimensional integrals givesApplying the same computation with reciprocal scale $1/t$ evaluates its inverse integral:For $f\in L^1$, Fubini's theorem gives $\widehat{f\ast g_t}=\hat{f}{e}^{-t|\xi {|}^2/2}$, which is integrable because $\hat{f}$ is bounded. Its inverse integral isAbsolute integrability of $|f(y)|e^{-t|\xi {|}^2/2}$ justifies this second application of Fubini's theorem. Thus Fourier inversion holds for every Gaussian convolution, in fact pointwise for this continuous representative.

Adopt the Boussinesq approximation, hydrostatic approximation, traditional approximation, and inviscid quasi-geostrophic approximation. Assume stable constant $N_0^2>0$, nonzero Coriolis parameter $f_0$, a beta plane, small Rossby number, $\beta L/|f_0|$ of that small order, and small displacement of the background stratification. Vertical advection of the buoyancy perturbation is then higher order. Define the quasi-geostrophic streamfunction byThe buoyancy and vertical vorticity equations at the retained order areDifferentiate the first in $z$. Since $J({\psi }_z,{\psi }_z)=0$, its material-derivative term becomes $f_0{D}_g{\psi }_{zz}$. Elimination of $w_z$ gives diabatically forced quasi-geostrophic potential vorticity:Thus it is the height gradient of heating, rather than heating alone, that produces potential vorticity at this order.

For the steady response linearized about rest, assume $\beta \ne 0$. The quasi-geostrophic potential-vorticity equation reduces to $\beta {\psi }_x=f_0{R}_z/N_0^2$. Choosing the horizontally uniform pressure gauge so that $\psi \to 0$ in the east gives the steady quasi-geostrophic response to localized heatingSetUsing the error function to integrate the Gaussian function givesThese velocities vanish as $x\to +\infty$. A depth-dependent but horizontally uniform addition to $\psi$ is dynamically irrelevant for the requested horizontal velocity. At $z=0$, streamlines are contours of $e^{-y^2/L^2}I(x)$. For $f_0,\beta ,R_0>0$, flow approaches the heating region from the west on its northern side, turns southward, and returns westward on its southern side. The contours are open, not closed gyres: $I$ tends to a nonzero constant in the west.

For $-H<z<0$, $R_z<0$ and the forcing decreases potential vorticity; steady balance requires $\beta v_g=f_0{R}_z/N_0^2<0$, giving southward flow. Below $z=-H$, $R_z>0$ and the meridional flow reverses. A constant heating rate independent of $z$ would have no such interior potential vorticity source.

For the nonlinear diagnostic quasi-geostrophic omega equation, write $\zeta ={\nabla }_h^2\psi$. Apply ${\nabla }_h^2$ to buoyancy evolution and $f_0{\partial }_z$ to vorticity evolution:Subtracting eliminates the pressure tendency and yieldsThe beta term is linear and must remain when the nonlinear terms are neglected. For the steady weak-forcing solution, $f_0\beta {\psi }_{xz}=f_0^2{R}_{zz}/N_0^2$, soThis also follows immediately from the steady linear buoyancy equation. It satisfies $w=0$ at $z=0$, decay at depth and horizontal infinity, and the linear quasi-geostrophic omega equation; with these homogeneous conditions the elliptic problem has no extra decaying homogeneous solution. The rigid-lid interpretation is an additional boundary approximation for this diagnostic response.

At $x=0$, the displayed $v_g$ and $w$ provide the requested meridional/vertical arrows. For positive heating there is upwelling concentrated near $y=0$, maximal in depth at $z=-H$, with northward flow below that depth and southward flow above. The vertical circulation tends to zero far from the forcing. A western zonal wake persists, but its meridional and vertical components vanish as $x\to -\infty$. Horizontal ageostrophic flow supplies the divergence associated with the upwelling; the projected $(v_g,w)$ slice is not itself a two-dimensional incompressible flow and need not have closed streamlines.

The steady calculation fixes an interior circulation, not its attainability from every possible initial and boundary state. For example, a rigid lid with initially uniform boundary buoyancy conserves that buoyancy at linear order because $R(0)=w(0)=0$. The formal steady profile instead generally has nonuniform $b^{\prime }(0)=f_0{\psi }_z(0)$. Such additional initial boundary data require a time-dependent adjustment and cannot simply be imposed on this particular steady solution. No boundary buoyancy data are supplied for the requested steady problem.

The ordinary saddle-point approximation with a nearby pole loses uniformity when the pole lies within the width of the saddle's Gaussian function profile. The distinguished limit isUse the exact quadratic phase coordinate $s=2i\sqrt{k}(u-1)$ and defineOn the descending contour, $s$ runs along the real axis from $-\infty$ to $+\infty$. The transformed differential has the useful partial fraction decompositionThe first term contains the nearby pole; the second is regular at the saddle and contributes $O(e^{-k}{k}^{-1/2})$. Approaching from below the descending contour means $\mathrm{Im}{s}_d<0$. The residue theorem then gives the Gaussian saddle-pole transitionThe exponent is $-s^2/4$, as in the original PDF; the converted TeX's $-s^2/2$ is a transcription error. The exact identity $k\varphi (z_0)=-k-s_d^2/4$ recovers the pole exponential in the requested expression.

The Gaussian pole integral expresses the bracket as $i\pi w(s_d/2)$, where $w$ is the Faddeeva function. Thus a convenient uniform leading answer isThe error estimate applies for bounded scaled pole position on the indicated side. In the overlap $1\ll |s_d|\ll \sqrt{k}$ below the real axis, the integral contributes $-2\sqrt{\pi }/s_d$ to leading order while the residue remains $2\pi i{e}^{-s_d^2/4}$. Since $s_d\sim \sqrt{k}(z_0-i)$, this reproduces the near-saddle limit of the separate saddle and residue terms. Boundary values on the descending contour are limits of this combined expression; the isolated real-axis pole integral itself needs a Cauchy principal value prescription.

For a normalized trial state $\psi$, expand it in an orthonormal basis of energy eigenstates, $\psi ={\sum }_n{c}_n{\psi }_n$. The Rayleigh quotient is thenThis is the quantum variational principle: minimizing the expectation of the Hamiltonian operator over any trial family gives an upper bound on the ground-state energy.

Let $\psi$ be the normalized ground-state wavefunction and form the normalized dilation ${\psi }_{\lambda }(x)={\lambda }^{1/2}\psi (\lambda x)$. Its kinetic and potential expectations areThe true ground state makes the energy stationary at $\lambda =1$, soThus the power-law quantum virial theorem gives

If $n\le -3$ and $A>0$, the two sides have opposite signs, because $T>0$ and $V>0$. If $A<0$, the scaled energy ${\lambda }^{2}T+{\lambda }^{-n}V$ tends to $-\infty$ as $\lambda \to \infty$, since $-n>2$, so the Hamiltonian is not bounded below. If $A=0$, the free particle has no normalizable ground state. Hence there is no normalizable ground state for $n\le -3$.

For $n=1$ take the normalized Gaussian functionThe Gaussian integral givesand thereforeIts unique minimum satisfiesAt this point $⟨V⟩=2⟨T⟩$, consistently with the virial theorem, and the resulting variational estimate is

The integrands are absolutely integrable because $f$ has compact support, $g\in L^1(\mathbb{R})$, and $|{\phi }_Z|\le 1$. Fubini's theorem therefore permits us to insert ${\phi }_Z(s)=\mathbb{E}[e^{isZ}]$ and interchange the expectation and integrals:With $r=\epsilon s$, the inner integral in $s$ isNow set $t=(Z-x)/\epsilon$. The change of variables $x=Z-\epsilon t$ giveswhich is the required identity.

Choose any $g$ with $g,\hat{g}\in L^1$, $g(0)\ne 0$, and for which Fourier inversion theorem applies, for example a Gaussian function. The left side factors through the Fourier transform of $f$ asOn the right side, $f(Z-\epsilon t)\to f(Z)$ andThe dominated convergence theorem and Fourier inversion at zero giveConsequently the Fourier-cutoff recovery of expectations from a characteristic function isThe right side depends only on ${\phi }_Z$. Equal characteristic functions therefore give equal expectations for every compactly supported continuous $f$, and part a yields the uniqueness theorem for characteristic functions.

The functional equation isEquivalently,

For a proof, letThe Poisson summation formula applied to a Gaussian function gives the theta transformationThe standard Gamma function integral and termwise integration initially give, for $\Re s>1$,Split the integral at one, substitute $u\mapsto 1/u$ in the lower half, and use the theta transformation. The result isThe integral is an entire function of $s$ because $\Theta (u)-1$ decays exponentially. The right side is visibly invariant under $s\mapsto 1-s$, proving both the analytic continuation and the functional equation of the Riemann zeta function. This is the Mellin representation of the completed Riemann zeta function.

The real Gaussian function heat kernel $(4\pi \epsilon z{)}^{-1/2}\exp [-(\varphi -\theta {)}^2/(4\epsilon z)]$ is integrable only when $\epsilon z>0$. For forward time $z>0$, this requires positive diffusivity. A Cole-Hopf transformation is an algebraic substitution for any nonzero coefficient, but that does not make the real forward convolution valid for negative diffusivity. Constant initial heat data already produce a divergent integral with the wrong sign.

Smooth forcing need not be integrable in phase space. For the hyperbolic characteristic flow for an inverted oscillator, the source $x^2+v^2$ accumulates as $q_t=\frac{1}{2}\sinh (2t)(x^2+v^2)-(\cosh (2t)-1)xv$. Its smallest quadratic-form eigenvalue is $(1-e^{-2t})/2>0$ for $t>0$. Starting with a nonnegative integrable Gaussian function, the solution is therefore unbounded and outside every finite Lp space at positive times. Locally time-integrable $L^p$ forcing prevents this failure by the Minkowski integral inequality.