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For fixed $c,h>0$, $L\ge 1$ and $\epsilon \ge 0$, complete the square in the exponent. A translated Gaussian function summed on a fixed-spaced lattice has mass $O(\sqrt{L})$ uniformly in the translate, by comparison with its integral and its maximum. This keeps a square-root logarithm, rather than the full number of dyadic intervals, when summing exponentially damped exponential sum bounds.

A Gaussian wave packet is a wavefunction whose spatial amplitude is a Gaussian function. Free quantum evolution preserves its Gaussian form while changing its complex width and spreading its probability density.

An even Dirichlet character satisfies $\chi (-1)=1$, while an odd Dirichlet character satisfies $\chi (-1)=-1$. Write $a=0$ or $1$ for its character parity and, for $x>0$, define the Dirichlet character theta functionFor a primitive Dirichlet character whose conductor of a Dirichlet character is $q>1$, the term at zero is zero. Put $\tau (\chi )={\sum }_{r\mathrm{mod}q}\chi (r)e(r/q)$, using the positive exponential, and ${\epsilon }_{\chi }=\tau (\chi )/(i^a\sqrt{q})$. The primitive Gauss sum of a Dirichlet character has magnitude $\sqrt{q}$, so $|{\epsilon }_{\chi }|=1$. The theta transformation isThus the powers are $x^{-1/2}$ in the even case and $x^{-3/2}$ in the odd case; the odd root number contains $1/i$. The conjugate character is necessary for a nonreal character. These formulas also follow by applying Poisson summation to the Gaussian function on each residue class, and to its derivative for odd parity. For the primitive principal Dirichlet character whose conductor of a Dirichlet character is one, use the ordinary Jacobi theta function with constant term one; its transformation has root number one.

The finite-$p$ assertion needs the zero-divergence hypothesis from the preceding part, which is not repeated in this part's printed hypotheses. The norm is on ${\mathbb{R}}_x^3\times {\mathbb{R}}_v^3$, and finite conservation requires $f_0$ to belong to the corresponding Lp space in addition to its smoothness.

Writing ${\Phi }_t$ for the characteristic flow map, the general solution is $f_t=f_0\circ {\Phi }_t^{-1}$. A change of variables givesThus the Lp conservation for incompressible transport isWithout that condition the requested conclusion is false: the force in part (c), together with any nonzero smooth integrable initial datum given by a Gaussian function, gives $\Vert f_t{\Vert }_p=e^{3t/p}\Vert f_0{\Vert }_p$. The essential supremum is still conserved by a complete invertible flow, because composition does not change the range of values.

Watson's lemma applies to Laplace-type endpoint integrals such as ${\int }_0^b{e}^{-\lambda t}f(t)dt$ as $\lambda \to +\infty$, when $f$ has a local algebraic expansion $f(t)\sim {\sum }_j{c}_j{t}^{p+j-1}$ with $p>0$ and suitable integrability or exponential-growth control away from the endpoint. Each term contributes $c_j\Gamma (p+j){\lambda }^{-p-j}$. The exponential selects a neighborhood of $t=0$.

Laplace method instead treats ${\int }_a^{b}g(t)e^{\lambda \varphi (t)}dt$ by locating the dominant maxima of a real phase $\varphi$. A smooth nondegenerate interior maximum produces a neighborhood governed by a Gaussian function of width ${\lambda }^{-1/2}$; endpoint or degenerate maxima require their corresponding local scaling. Contributions away from a strict dominant maximum must be smaller. These methods overlap after suitable changes of variable, but their usual starting forms emphasize an endpoint amplitude expansion and an exponential maximum, respectively.

For the Gamma function, put $t=xu$. ThenThe phase has its unique maximum at $u=1$, with value $-1$ and second derivative $-1$. Set $u=1+s/\sqrt{x}$. Near this maximum,Combining the amplitude and exponential expansions in Laplace method givesLocalization near $u=1$ justifies extending the limits for the Gaussian integral; the distant endpoint is not being expanded uniformly by this local series. Odd terms integrate to zero. The moments of the Gaussian integral satisfy $I_1=I_0$, $I_2=3I_0$, $I_3=15I_0$, so the relative correction isThe two terms of the Stirling formula, obtained by the Stirling correction by Gaussian moments, are therefore

Here the singular perturbation has an exact conservative structure:The integrating factor for this first-order linear differential equation givesAt $x=1$ and $x=-1$ the integral terms have opposite signs while the exponential factors agree. Equal boundary data therefore force $C=0$, and normalization gives the unique exact solutionThis exposes the Gaussian interior layer of a conservative drift equation. Its central scale is $x=O(\sqrt{ϵ})$, where $x=\sqrt{ϵ}\xi$ gives $Y^{\prime \prime }+\xi Y^{\prime }+Y=0$. The central Gaussian function profile is $e^{-{\xi }^2/2}$ with amplitude $e^{1/(2ϵ)}$.

Away from the center, $|x|\gg \sqrt{ϵ}$, use exponential WKB approximation branches rather than an assumed bounded algebraic outer solution. The reduced ordinary differential equation $x{y}_0^{\prime }+y_0=0$ would give $C/x$; matching both equal positive endpoint values with that algebraic branch is impossible. The exact solution selects its zero coefficient and the exponentially large branch with a Gaussian function profile instead. Near either endpoint, a distance $O(ϵ)$ changes $y$ by order one relative to its endpoint value: with $r=(1-|x|)/ϵ$, $y=e^{r-ϵr^2/2}$. These endpoint scales describe normalization of the same global exponential solution.

Thus a central $O(\sqrt{ϵ})$ turning region, exponential outer regions on both sides, and $O(ϵ)$ endpoint normalization scales give the complete regional description. The peak is exponentially large, so no uniformly bounded regular outer expansion should be presumed. The exact expression already provides a uniform solution without further matching calculations.

Set $q=\log \psi$ and use the sign-appropriate Cole-Hopf transformation $f=2\epsilon q_{\theta }$. Direct differentiation givesThus the heat equation ${\psi }_z=\epsilon {\psi }_{\theta \theta }$ implies the viscous Burgers equation with the required negative nonlinear sign. Conversely, vanishing of the Burgers residual makes the final quotient depend only on $z$, and a multiplicative time-dependent factor in $\psi$ removes it. The algebraic substitution works for any nonzero $\epsilon$ where $\psi \ne 0$.

For a forward dissipative initial-value solution, take $\epsilon >0$ and $z>0$. This is essential for the supplied heat kernel: at $\epsilon <0$, $z>0$, its real Gaussian integral diverges even for $\psi (0,\varphi )=1$. More generally that kernel requires $\epsilon z>0$. The physical viscosity convention selects the positive case; the mere condition $\epsilon \ne 0$ does not justify a forward Gaussian function formula.

Integrating ${\partial }_{\theta }\log \psi (0,\theta )=f_0(\theta )/(2\epsilon )$ gives a positive continuous initial factor, normalized asIts heat kernel convolution converges for both signs of $U$, since a Gaussian function dominates the one-sided exponential. DefineThe left half-line contributes $J(\theta )$; completing the square on the right half-line gives $e^{S}J(-\theta -Uz)$. ThusWhen differentiating, the two moving-endpoint Gaussian function terms cancel, becauseOnly ${\partial }_{\theta }{e}^S=(U/(2\epsilon ))e^S$ remains in the logarithmic derivative. The viscous Burgers step solution with negative flux is thereforeThe denominator is strictly positive, so $f$ lies between $0$ and $U$, irrespective of the sign of $U$. Its initial limits away from $\theta =0$ are the required step.

For the spatial limits, the Gaussian function tails must be compared with $e^S$; inspecting that exponential alone gives the wrong inference when $U<0$. As $b\to +\infty$, the complementary error function asymptotic givesAt $\theta \to +\infty$, $J(\theta )e^{-S}$ tends to zero even for $U<0$, because its quadratic decay dominates any exponential linear in $\theta$. Hence $\alpha e^{-S}\to 0$. At $\theta \to -\infty$, use the identity above to obtainwhile $J(\theta )$ approaches its full Gaussian integral. Thusfor both signs of $U$.

Since $J$ is strictly decreasing, $\alpha =1$ holds exactly when $\theta =-\theta -Uz$. At this point $S=0$, so the midpoint symmetry of a viscous Burgers step givesMore generally the exact symmetry is $f(z,-Uz-\theta )=U-f(z,\theta )$. For $U>0$, the midpoint follows the inviscid shock trajectory. For $U<0$, it is instead the center of an expanding fan, not a shock.

The vanishing viscosity approximation makes the comparison precise. For fixed $z>0$ and $U>0$, near $\theta =-Uz/2$ both tail integrals tend to the full Gaussian integral, so $\alpha \to 1$ andThe layer has width $O(\epsilon /U)$ and tends to the entropy shock, with value $U/2$ at its center. Outside it the limits are $0$ and $U$.

For $U<0$, inside $0<\theta <-Uz$ both Gaussian function tails have large positive lower limits. Their asymptotics giveOutside this interval the limits are $0$ on the left and $U$ on the right. Thus positive viscosity selects the rarefaction wave, including the same midpoint value as the inviscid fan. The two sign cases agree with the preceding entropy solution construction, while a negative diffusivity would not furnish this dissipative selection.