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The Engquist-Osher method is a conservative finite volume method built by separating positive and negative characteristic speeds. Take $f\in C^1$, a uniform cell width $d$, a time step $k$, and cell averages $U_j^n$. Write $\lambda =k/d$ and define the numerical fluxThe Engquist-Osher flux uses the left state for right-going transport and the right state for left-going transport. Its conservative update isIt is consistent because $F(u,u)=f(u)$. The integrals have their ordinary oriented meaning even for negative states. For the Inviscid Burgers equation, $f(u)=u^2/2$ gives the useful exampleIn a region where all characteristic speeds have one sign, the flux reduces to the corresponding upwind flux. In smooth regions, this basic piecewise-constant, forward-time version is first order in space and time; its main virtue is robust nonlinear stability across shocks.

Here is a precise stability proof. Assume the data lie in a bounded interval $J=[m,M]$ and chooseWork on a periodic grid, or on the infinite grid with summable differences; on a finite interval the inflow boundary data and boundary fluxes must be treated monotonically as well. Define the three-point update mapWriting $f_{+}^{\prime }=\max (f^{\prime },0)$ and $f_{-}^{\prime }=\min (f^{\prime },0)$, its three partial derivatives areThus it is a monotone conservative scheme. Since $H(s,s,s)=s$, comparison with the constant states provesThe interval is invariant, so the same Courant–Friedrichs–Lewy condition remains valid at every step. This is an $L^{\infty }$ bound relative to constant states, not a claim that arbitrary pairs of solutions are contractive in $L^{\infty }$.

For a stronger perturbation estimate, let $\mathcal{T}$ denote the global update and use componentwise maxima. Monotonicity gives $\mathcal{T}(U\vee V)\ge \mathcal{T}U,\mathcal{T}V$. ThereforeSum over the grid. The shared interface flux cancels telescopically, so conservation givesInterchanging $U,V$ and adding yields the L1 contraction of a monotone conservative scheme:This is a mesh-independent nonlinear stability estimate. On an infinite grid the telescoping argument is justified by summability of the differences and the bounded characteristic-speed range, or by truncation followed by a limit. For unequal prescribed boundary data, additional boundary flux terms enter the estimate.

The method is also a total variation diminishing scheme. Let $\mathcal{S}$ shift a grid sequence by one cell. Translation invariance gives $\mathcal{T}\mathcal{S}=\mathcal{S}\mathcal{T}$. Applying the preceding $L^1$ contraction to $U$ and $\mathcal{S}U$ showsHence initially finite discrete total variation cannot grow.

Finally, monotonicity supplies a discrete entropy inequality, explaining why the stability is compatible with the physically admissible entropy solution. For a constant $c\in J$, defineComparison of $H$ with the clipped triples above and below $c$ givesIndeed $H(U\vee c)\ge \max (H(U),c)$ and $H(U\wedge c)\le \min (H(U),c)$, and subtracting these two update formulas yields exactly the right side. Also $Q(u,u;c)=\mathrm{sgn}(u-c)(f(u)-f(c))$, the Kruzhkov entropy flux. Under the displayed CFL condition the method therefore has an invariant state range, $L^1$ contraction, decreasing discrete total variation and entropy dissipation. These estimates remain meaningful at discontinuities, where linearized Fourier analysis alone cannot establish the corresponding nonlinear stability.

For this non-Archimedean local field, the Schwartz-Bruhat space $\mathcal{S}(F)$ consists of locally constant, compactly supported complex-valued functions. Choose a nontrivial continuous additive character $\psi :F\to {\mathbb{C}}^{\times }$ of modulus one and an additive Haar measure $dx$. We use the plus-sign convention for the Fourier transform over a local field:Compact support makes the integral absolutely convergent. The additive character and the scale of the Haar measure are part of the definition; without them there is no canonical numerical transform.

Let $q=|{\mathcal{O}}_F/{\pi }_F{\mathcal{O}}_F|$, and write $V=\mathrm{vol}({\mathcal{O}}_F)$. Define the integer $c$ by the annihilator of the valuation ring ${\mathcal{O}}_F^{\perp }={\pi }_F^c{\mathcal{O}}_F$. Equivalently, $\psi$ is trivial on ${\pi }_F^c{\mathcal{O}}_F$ and not on ${\pi }_F^{c-1}{\mathcal{O}}_F$. Such a conductor exists: continuity puts the image of some additive ball inside an arc containing no nontrivial circle subgroup, so the character is trivial on that ball; nontriviality bounds the possible ball exponents below. Then $H_n={\pi }_F^n{\mathcal{O}}_F$ has volume $V{q}^{-n}$ and character annihilator $H_n^{\perp }={\pi }_F^{c-n}{\mathcal{O}}_F$.

Translation givesThe integral is the volume if $y\in H_n^{\perp }$. Otherwise translate it by $t_0\in H_n$ with $\psi (t_{0}y)\ne 1$; Haar measure invariance multiplies the same integral by a nonidentity scalar, so it must vanish. ThusEvery Schwartz-Bruhat function is a finite linear combination of such coset indicators: compactness of its support supplies a common sufficiently small translation subgroup on which it is constant, and finitely many of its cosets cover the support. The transform of each indicator has compact support and is locally constant, since $\psi$ has open kernel. Consequently $\hat{f}\in \mathcal{S}(F)$ for every $f\in \mathcal{S}(F)$.

Apply the transform again to an indicator. The same character-orthogonality calculation, together with $(H_n^{\perp }{)}^{\perp }=H_n$, givesChoose the self-dual Haar measure, characterized here by $V=q^{c/2}$. By linearity the desired Fourier inversion isIn particular, one may rescale any nontrivial additive character to make $c=0$, and then take $V=1$. For a concrete construction, start with ${\psi }_0(x)={\psi }_{{\mathbb{Q}}_p}({\mathrm{Tr}}_{F/{\mathbb{Q}}_p}x)$, where the standard rational additive character has kernel ${\mathbb{Z}}_p$. Its annihilator is the inverse different ${\mathfrak{D}}_{F/{\mathbb{Q}}_p}^{-1}$. If that ideal is ${\pi }_F^{-d}{\mathcal{O}}_F$, the rescaled character $\psi (x)={\psi }_0({\pi }_F^{-d}x)$ has ${\mathcal{O}}_F^{\perp }={\mathcal{O}}_F$. This supplies the stated normalization and also explains how the different ideal enters local Fourier analysis.