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Particles keep their velocity and move along straight trajectories. The method of characteristics gives $f(t,x,v)=f_0(x-tv,v)$. The shear preserves Lebesgue measure and every Lp norm. Compact initial support remains compact at each finite time, but need not remain in one compact set for all times.

For the free transport equation, the mixed Lebesgue norm dispersion estimate isIndeed, use the method of characteristics and then the change of variables $y=x-tv$:Taking the essential supremum over $x$ proves the result. This decay measures the spreading of a velocity average; the total phase space Lp norm is conserved rather than decaying. The datum on the right is the initial datum, not the solution at the same time.

An oscillatory integral defines a distribution by cancellation, even when its oscillatory integral amplitude is not integrable in frequency. We use symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)=S_{1,0}^N(X\times {\mathbb{R}}^k)$, where $X\subset {\mathbb{R}}^d$ is open and $N\in \mathbb{R}$. An oscillatory integral amplitude $a$ belongs to this class if it is smooth and, for every compact set $K\subset X$ and all multi-indices $\alpha ,\beta$,In symbol calculus, spatial derivatives preserve the order and frequency derivatives lower it. The fixed number $N$ is independent of $K$; this will produce a finite global order of a distribution, although continuity constants may depend on $K$.

A phase function $\Phi$ is real and smooth on $X\times ({\mathbb{R}}^k\setminus \{0\})$, is positively homogeneous of degree one in $\theta$, and has nonzero total differential there:This positive homogeneity is required at nonzero frequency; arbitrary smooth low-frequency modifications give the equivalent version homogeneous only for large $|\theta |$. In particular, smoothness at zero is not an additional requirement on a general homogeneous phase function. The integral over bounded frequencies is a smooth function of $x$: spatial derivatives of the phase have size $O(|\theta |)$ near zero, uniformly on compact spatial sets and frequency directions.

Choose a cutoff function $\chi \in C_c^{\infty }({\mathbb{R}}^k)$ equal to one near zero. The proposed meaning of the oscillatory integral distribution isExistence and cutoff independence of an oscillatory integral require a proof. Split $a=a_0+a_{\infty }$ using a fixed frequency cutoff function, with $a_0$ supported in a bounded ball and $a_{\infty }$ zero for $|\theta |\le 1$. The low-frequency part is already absolutely integrable. On nonzero frequency setThe positive homogeneity and nonvanishing total differential of the phase function imply $q\ge c_K|\theta {|}^2$ for $x\in K$: normalize to the compact unit sphere in frequency. Direct differentiation gives $L{e}^{i\Phi }=e^{i\Phi }$.

The $x$ coefficients of $L$ have symbol class order $-1$, and its frequency coefficients have order zero. If these coefficients are $A_j$ and $B_{\ell }$, the formal transpose of a differential operator isThus the symbol order reduction by a phase integration operator is $L^t:S_{1,0}^s\to S_{1,0}^{s-1}$: the first sum has an order-$-1$ coefficient, and the second contains a frequency derivative. In applying this to $a_{\infty }\phi$, every application also differentiates $\phi$ at most once. Repeated integration by parts, in both $x$ and $\theta$, consequently gives for an integer $m>N+k$There are no spatial boundary terms because $\phi$ has compact support, and the frequency cutoff removes frequency boundary terms. Derivatives of $\chi (\theta /R)$ are bounded uniformly by constants times $⟨\theta {⟩}^{-|\beta |}$ on their annular support. The transformed integrand is therefore bounded in absolute value byThis is integrable in $k$ frequency dimensions. Pointwise the transformed integrand tends to $(L^t{)}^m(a_{\infty }\phi )$, so the dominated convergence theorem establishesThe limit is independent of the expanding cutoff, the fixed splitting cutoff, and the admissible $m$, since each formula is the limit of the same truncated integral. Derivatives landing on the expanding cutoff can also be estimated directly by $O(R^{N-m+k})$, which tends to zero. In particular,This proves continuity on the space of test functions and the finite order of an oscillatory integral distribution. The same $m$ works for all compact sets. When $N<-k$, $m=0$ is possible and the original frequency integral is absolutely integrable; cancellation is needed for general $N$.

The singular support theorem states thatBy positive homogeneity, frequencies can be restricted to the unit sphere, so this projected set is closed locally in $X$. More precisely, the stationary-direction bound for singular support restricts the frequency directions to the closed conic support of an oscillatory amplitude; using a closed directional support avoids losing limits occurring at arbitrarily high frequency. The simpler displayed bound is sufficient here.

To prove the bound, take $x_0$ outside the displayed stationary set. On a sufficiently small spatial neighborhood and all unit frequency directions, $|{\nabla }_{\theta }\Phi |$ is bounded below. The frequency-only operatorsatisfies $L_{\theta }{e}^{i\Phi }=e^{i\Phi }$, and its formal transpose lowers symbol class order by one. A spatial derivative of order $|\alpha |$ of $e^{i\Phi }a$ has amplitude order at most $N+|\alpha |$. Applying the frequency integration by parts more than $N+|\alpha |+k$ times makes that derivative absolutely integrable, uniformly on smaller compact sets. Every spatial derivative therefore exists and is continuous there. The oscillatory integral distribution is a smooth function near $x_0$, proving the singular support assertion.

For the linear transport equation, use spacetime $X={\mathbb{R}}^n\times (0,\infty )$, frequency $\theta \in {\mathbb{R}}^n$, andThis is a valid phase function, because ${\nabla }_x\Phi =\theta \ne 0$ at nonzero frequency; the oscillatory integral amplitude is in symbol class order zero. The Fourier representation of the Dirac delta function gives the transport of a Dirac point mass:Its rigorous spacetime distribution pairing is $⟨u,\phi ⟩={\int }_0^{\infty }\phi (ct,t)dt$. Consequentlywhere the endpoints vanish since a spacetime test function has compact support in $t>0$. At each fixed $t$, $⟨u(t),\psi ⟩=\psi (ct)$, so weak convergence of distributions gives $u(t)\to {\delta }_0$ as $t\downarrow 0$. This is the required initial trace.

Finally, ${\nabla }_{\theta }\Phi =x-ct$, so the singular support theorem confines singularities to $x=ct$. In fact equality holds: $u$ is a nonzero order-zero distribution on that trajectory and vanishes off it. A smooth function supported on this set of empty interior must vanish, so $u$ cannot be smooth in any neighborhood of a point of the trajectory. ThusThe method of characteristics also proves uniqueness among solutions with a distributional initial trace: the change of variables $y=x-ct$ turns the linear transport equation into ${\partial }_{t}w=0$. Such a distribution is constant in $t$; pairing with spatial test functions reduces this assertion to a distribution with zero derivative is constant. Its initial trace fixes $w={\delta }_0$, so the moving Dirac delta distribution above is the unique solution. There is transport of the singularity along the characteristic and no smoothing.

Take $x$ increasing upslope and let $u$ be the superficial Darcy velocity, so the mobile discharge is $uh$. Its pore transport speed is $u/\varphi$. This interpretation is required by the factors of porosity in the printed equations; a literal pore-speed convention would instead replace $u$ there by $\varphi u$.

For capillary residual trapping, let $M(x,t)={\max }_{0\le \tau \le t}h(x,\tau )$ be the maximum invaded thickness, including the initial state. The stored carbon-dioxide volume per plan area isDuring advance, $M=h$ and new pores fill with mobile carbon dioxide; during recession, $M$ is fixed and the newly vacated pores retain saturation $s$. Conservation of mobile plus trapped fluid isConsequentlyAssume $a,L,u,\varphi >0$ and $0<s<1$. The two speeds differ because the advancing front fills a whole pore volume while the receding tail removes only its mobile fraction.

The method of characteristics keeps height constant on $x=x_0+v_{A}t$ on the leading face and $x=x_0+v_{R}t$ on the trailing face. Matching the two linear profiles gives the triangular current with capillary retention:Before extinction, the rear, front, crest position and crest height areEvery positive height has one trailing and one leading characteristic; their intersections trace the crest. The mobile current disappears at

The final trapped region is the maximum-invasion envelope of a retained current. For $0<x<L$, the initial height is already the maximum, $M_{\infty }=ax$. For $L<x<x_{\ast }$, the maximum occurs when the crest passes $x$, at $t_m=2(x-L)/(v_R+v_A)$. Substituting into the trailing profile givesIf $z$ is distance into the aquifer measured normally from its upper boundary, trapped carbon dioxide occupies $0<z<M_{\infty }(x)$ at residual pore saturation $s$. The occupied pore volume per unit transverse width isexactly the initial mobile volume. This is an independent mass-conservation check of both the envelope and extinction distance. In the limit $s=0$, the current translates without shrinking and nothing is trapped; the finite-extinction formula is not used in that limit.

For the negative-flux Inviscid Burgers equation, the method of characteristics givesThe characteristic curve starting at ${\theta }_0$ therefore has $\theta ={\theta }_0-f_0({\theta }_0)z$, and henceFor smooth data this describes a single-valued classical solution as long as the characteristic flow map is invertible. Its Jacobian is $1-z{f}_0^{\prime }({\theta }_0)$; after characteristic crossing, one must instead select a weak solution with the appropriate entropy solution condition.

The conservation form is $f_z+G(f{)}_{\theta }=0$, with conservation law flux $G(f)=-f^2/2$. Integrating this scalar conservation law across a moving discontinuity, or differentiating its Heaviside representation as a distribution, yields the Rankine-Hugoniot conditionFor distinct one-sided limits it simplifies toThis jump-speed relation is exact for the Burgers conservation law; an additional entropy condition is needed to distinguish a physical compressive shock wave from an expansion discontinuity.

The step gives the Burgers Riemann problem with negative flux. If $U>0$, characteristic curves from the left have speed zero, while those from the right have speed $-U$: they converge. The entropy shock wave has speed $-U/2$ and thusCharacteristics enter this shock from both sides, since $0>-U/2>-U$.

If $U<0$, the right-hand speed $-U$ is positive and the two families separate. A smooth steep approximation to the initial step spreads into a rarefaction wave. In the fan, the self-similar characteristic curve relation is $-f=\theta /z$, givingThe endpoint values match continuously. The discontinuity moving at $-U/2$ would satisfy the jump condition even for $U<0$, but its characteristic curves leave the discontinuity and it fails entropy admissibility. For $U=0$ the solution is identically zero.

For the sharp energy estimate, $C_0=0$ stays uniform while the available bound $C_1=M^2/\epsilon$ diverges as $\epsilon \downarrow 0$ when $M>0$. If the coarser estimate is used for the first part, both displayed bounds $M^2/\epsilon$ diverge, but the first divergence is only an artifact of discarding an exact cancellation.

The method of characteristics explains why uniform control of the gradient cannot generally persist for the inviscid scalar conservation law. Before characteristic crossing, with initial data $u_0$,If $F^{\prime \prime }(u_0(\xi ))u_0^{\prime }(\xi )<0$ somewhere, the denominator reaches zero in finite positive time. The solution steepens and the smooth description breaks down; an entropy solution can subsequently contain shocks. Positive viscosity replaces such a discontinuity by a thin smooth layer, which can have large gradient even while its $L^2$ norm remains controlled. The divergent bound does not assert that every flux and every initial datum form a shock; a linear flux, for example, has no such steepening.

Write the negative-flux Inviscid Burgers equation as $f_Z+{\partial }_{\theta }F(f)=0$, with $F(f)=-f^2/2$. Along a characteristic curve,so a characteristic starting at ${\theta }_0$ has $\theta ={\theta }_0-f_0({\theta }_0)Z$ andThis formula is a classical solution only while the method of characteristics is one-to-one; after characteristic crossing one must select an entropy solution.

For a discontinuity with left and right values $f_{-}$ and $f_{+}$, the Rankine-Hugoniot condition follows by integrating conservation across a moving small interval:For a nonzero jump, factor the difference of squares to obtainThe negative flux makes increasing jumps compressive. Decreasing jumps spread into rarefaction waves; they cannot be retained as nonphysical expansion shock waves.

In the central ramp of each period, $f_0({\theta }_0)={\theta }_0-2m$, so its method of characteristics is $\theta -2m=(1-Z)({\theta }_0-2m)$. For $0<Z<1$, the retained ramp therefore occupies $|\theta -2m|<1-Z$. At the odd boundary $r=2m+1$, the initial limiting values are $+1$ on the left and $-1$ on the right. Their characteristic speeds are $-1$ and $+1$, giving the fan $f=-(\theta -r)/Z$ on $|\theta -r|<Z$. Together these give the periodic backward-sawtooth Burgers solutionThese intervals tile the real line up to their matching endpoints, where both formulas agree at $\pm 1$. The increasing ramp steepens, but the wave's maximum magnitude remains one before breaking. A steep continuous regularization of the original downward jump produces exactly the limiting fan, as suggested by the characteristic construction.

At $Z=1$, each increasing ramp collapses at $\theta =2m$. The fans on either side meet there with values $-1$ and $+1$, producing a compressive stationary shock wave. For all $Z\ge 1$ the fan between successive shock waves remains centered at the odd point, givingAt a shock wave the limiting values are $f_{-}=-1/Z$ and $f_{+}=1/Z$. Their average is zero, so the Rankine-Hugoniot condition keeps the shock wave fixed. Their characteristic speeds satisfy $1/Z>0>-1/Z$, confirming compression into the shock wave. The arbitrary pointwise value at a shock wave does not affect the weak solution. The post-breaking amplitude decays as $1/Z$, despite the lack of explicit viscosity, because shock wave dissipate the wave.

For the requested sketches, $Z=1/3$ has ramp slope $3/2$ on $|\theta -2m|<2/3$ and fan slope $-3$ around odd points on width $2/3$. It is still continuous. At $Z=3$, the profile decreases linearly with slope $-1/3$ between even points and jumps from $-1/3$ to $+1/3$ at each even point: