Transport equation ($u_t+b\cdot \nabla u=0$)

A transport equation carries a quantity along the flow of a velocity field $b$. For a smooth solution of $u_t+b\cdot \nabla u=0$, the value of $u$ is constant along each characteristic curve. With constant speed $c$, the chain rule gives $\frac{d}{dt}u(x_0+ct,t)=u_t+c{u}_x=0$, hence $u(x,t)=g(x-ct)$. This is exact translation: it preserves the shape rather than smoothing it, as the periodic heat and transport comparison shows.

Suppose $D=a{\partial }_x+b{\partial }_y$, $DA=f$, $Dq=0$, and $A,q$ are smooth functions. Then $u=A+g(q)$ solves $Du=f$ wherever $g$ is differentiable. If $g$ is continuous and piecewise continuously differentiable with a jump in $g^{\prime }$ at $q=q_0$, the one-sided gradients of $u$ differ by $[g^{\prime }]\nabla q$ across that characteristic curve. Thus a nonzero derivative jump in the data persists along the characteristic unless the relevant component of $\nabla q$ vanishes. A locally Lipschitz continuous $g$ also gives a weak solution across the curve.

Use dimensionless $x$ and $t$, the periodic domain $[0,2\pi ]$, and the same initial conditionWith zero PDE source term, compare the heat equation $u_t=u_{xx}$ and the transport equation $u_t+u_x=0$. Their exact solutions areEach cosine satisfies ${\partial }_x^2\cos (nx)=-n^2\cos (nx)$, so the heat equation reduces its amplitude according to $A_n^{\prime }=-n^2{A}_n$. The superposition principle combines the constant and two Fourier modes. The transport equation shifts the argument of the entire profile. Both formulas meet the initial condition and matching periodic boundary conditions exactly.^[ref]

At $t=1$, the heat amplitudes are approximately $0.368$ and $0.0000494$: the small ripples disappear much sooner than the broad variation. The transport equation retains amplitudes $1$ and $0.4$, shifting the profile right by one unit with wraparound. Both preserve spatial average $1$, but only the heat equation approaches the uniform profile. This gives diffusion its usefulness for smoothing uneven temperatures, while the transport equation models a profile carried by a flow. The signed profile can represent a scaled temperature deviation.

To restore units, let physical position be $X=\ell x$ and time be $T=\tau t$. For $U_T=D{U}_{XX}$, the dimensionless coefficient is $D\tau /{\ell }^2$, with $D$ measured in length squared per time. For $U_T+c{U}_X=0$, it is $c\tau /\ell$, with $c$ measured in length per time. Both plotted coefficients equal one when $D={\ell }^2/\tau$ and $c=\ell /\tau$. Thus the example compares specified models on the same scales, rather than identifying diffusivity with velocity.

Let $a=\mu (1+\mu )/2$, $b=(1+\mu )(2-\mu )$, $c=(1-\mu )(2-\mu )/2$, $d=2-\mu$, and $e=1+\mu$. The implicit stencil $a{U}_{j-1}^{n+1}+b{U}_j^{n+1}+c{U}_{j+1}^{n+1}=d{U}_j^n+e{U}_{j+1}^n$ for $u_t=u_x$ has centered characteristic moments zero through degree three and fourth moment $K(\mu )$. Its Fourier numerator and denominator satisfy $|L(\theta ){|}^2-|N(\theta ){|}^2=K(\mu )(1-\cos \theta {)}^2$. The marching operator is a contraction exactly for nonnegative Courant numbers in $[0,1]\cup [2,\infty )$; with signed Courant numbers, also include $(-\infty ,-1]$. At the four roots of $K$ the method becomes an exact integer grid translation rather than merely third order.

Transport along a Hamiltonian flow. The solution is $f(\cdot ,t)=f_I\circ {\Phi }_{-t}$ wherever the inverse characteristic exists. Equality of mixed derivatives makes the transport vector field divergence-free, so its flow preserves phase-space volume. Smoothness of the Hamiltonian alone gives local characteristics, not a complete flow on all of phase space.

For an incompressible transport flow, every integrable function of the transported scalar has constant integral: the scalar is unchanged along characteristics and the flow's Jacobian determinant equals one. For a finite integral on infinite phase space, require integrability; $F(0)=0$ and compactly supported data are one common choice. For a whole-space conservation statement also require no escape or boundary flux at infinity.

The differential operators $Z_j$ commute with $D={\partial }_t+a(v)\cdot {\nabla }_x$. Indeed $[D,{\partial }_{v_j}]=-{\sum }_i({\partial }_{v_j}{a}_i){\partial }_{x_i}$, canceled by differentiating the explicit time factor in $Z_j$. Hence $Z_{j}f$ satisfies the same transport equation as $f$, and obeys the same L2 norm conservation when boundary fluxes vanish. For the standard Jacobian matrix $(Da{)}_{ij}={\partial }_{v_j}{a}_i$, the vector of these derivatives is ${\nabla }_{v}f+t(Da{)}^T{\nabla }_{x}f$.

For $f(t,x,v)=f_0(x-ta(v),v)$, apply the area formula to the velocity map and bound the initial data by their velocity essential supremum. This gives the displayed mixed Lebesgue norm estimate whenever the weighted inverse multiplicity is essentially bounded. If $a$ is injective and $|\det Da|\ge \alpha >0$, the constant is at most $1/\alpha$. In dimension one a derivative bounded away from zero gives a global diffeomorphism; in higher dimensions determinant bounds alone do not.

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, substitute $y=x-tv$ into $\int |f_0(x-tv,v)|dv$. The resulting factor $|t{|}^{-d}$ and the bound by ${\sup }_v|f_0(y,v)|$ prove this mixed Lebesgue norm estimate. The right side contains initial data. The same-time mixed norm is not a conserved replacement: prescribing a product $\psi (x)\chi (v)$ at a later time and dilating $\chi$ disproves such a claim.

A distributional weak solution of ${\partial }_{t}f+v\cdot {\nabla }_{x}f=g$, with locally integrable $f,g$, becomes ${\partial }_{t}F=G$ after the shear $F(t,x,v)=f(t,x+tv,v)$. For almost every characteristic label, there is an absolutely continuous function representing $F$, and its increments are the integrals of $G$ for every pair of times. Arbitrary modifications on time slices can destroy this every-time identity without changing the distributional solution.

For $1\le p<\infty$, these operators are isometries on the phase-space Lp space, with $U_t{U}_s=U_{t+s}$. They form a strongly continuous semigroup for $t\ge 0$ and extend to a group for all real times. Strong continuity follows by smooth compactly supported approximation and measure preservation. They preserve the essential supremum as well, but strong continuity is not claimed on the whole $L^{\infty }$ space.

For the transport field $b(t,x)$, a characteristic curve solvesThe chain rule converts differentiation of $u(t,X(t))$ into the transport operator applied to $u$.

The radial transport equation has projected characteristic curves $(x,y)=e^s(x_0,y_0)$, which remain on positive rays. Data on $x+y=1$ reach precisely the half-plane $x+y>0$, since $x+y=e^s$. A datum $u(a,1-a)=h(a)$ therefore gives $u(x,y)=(x+y{)}^{k}h(x/(x+y))$ in that half-plane. The opposite half-plane is not determined by these data. Extending a characteristic line through the origin would join distinct flow orbits across a point where the projected vector field vanishes.

For $a{u}_x+b{u}_y=c$, the projected vector field $(a,b)$ has the characteristic projections as its integral curves. The lifted vector field is $(a,b,c)$ in $(x,y,u)$ coordinates. A vanishing projected vector field can obstruct a classical solution even when the lifted vector field does not vanish.

For the first-order partial differential equation $a(x,y,z)z_x+b(x,y,z)z_y=c(x,y,z)$, a lifted characteristic curve satisfies $(x^{\prime },y^{\prime },z^{\prime })=(a,b,c)$. Its characteristic projection is the image $(x(s),y(s))$ in the plane of independent variables. The projection need not be a regular curve: if $a=b=0$, it can be stationary while $z$ varies. In that situation the lifted curve cannot lie in the graph of a single-valued classical solution unless the required compatibility holds. Noncharacteristic Cauchy data require the tangent to the initial curve and the projected characteristic vector to be linearly independent.

The characteristic flow map $Z_{s,t}(x)$ is the point reached at time $t$ by the characteristic curve that passes through $x$ at time $s$. Whenever the flow is invertible, transport solutions pull their initial data back by $Z_{0,t}^{-1}$.

Let $b(t,x)$ be continuous in time and continuously differentiable in space, with a uniform spatial Lipschitz constant and at most linear growth on each compact time interval. Its characteristic flow map exists forward and backward over every bounded time interval. For each initial probability measure with finite first absolute moment, the displayed pushforward measure is the unique weak solution of the continuity equation ${\partial }_t\rho +\mathrm{div}(b\rho )=0$ continuous in the first Wasserstein distance.

The Gronwall inequality bounds the flow by $C_T(1+|x|)$ and therefore preserves finite first absolute moments. The chain rule establishes the weak equation. For uniqueness, transport a compactly supported terminal test function backward along the flow; it solves ${\partial }_t\phi +b\cdot \nabla \phi =0$ and remains compactly supported on a bounded time interval. Inserting it into the weak identity identifies every terminal integral with that of the claimed pushforward measure. A zero divergence makes the corresponding scalar transport equation and the continuity equation coincide.

The Jacobian determinant of a differentiable characteristic flow map measures its change of volume. The variational equation and the Liouville formula for a fundamental matrix give the displayed formula for a complete smooth field $b$. For kinetic advection with $b=(v,F)$, its divergence is ${\nabla }_v\cdot F$. A zero divergence preserves phase-space Lebesgue measure.

If $F\in C^1(\mathbb{R}\times {\mathbb{R}}^d)$ and $|F(t,x)|\le K(1+|x|)$, every characteristic curve is defined for all real time. The Gronwall inequality bounds $1+|X(t)|$ on each bounded time interval, so local ordinary differential equation existence cannot terminate through escape to infinity. Uniqueness and the variational equation give a global $C^1$ spatial flow diffeomorphism, with Jacobian $J>0$ satisfying $J_t=(\mathrm{div}F)(t,X)J$.

The method of characteristics solves a first-order partial differential equation by reducing it to ordinary differential equations along characteristic curves.

For the partial differential equation $u_x+u{u}_y=x$ with initial profile $u(0,\eta )=1+\sin \eta$, the method of characteristics gives the parametrized surfaceThe projection to the $(x,y)$ plane has Jacobian determinant $1+x\cos \eta$. It is positive for $|x|<1$, so the inverse function theorem makes the surface a unique local solution graph there. For $\eta \ge 0$, strict monotonicity and divergence to infinity give a global inverse onto $y\ge x+x^3/6$ at each such $x$.

The derivative $u_y=\cos \eta /(1+x\cos \eta )$ becomes singular when that projection first degenerates, although $u$ itself stays finite. At $x=1$ the degenerate labels are odd multiples of $\pi$, and at $x=-1$ they are even multiples. At either boundary the limiting map remains strictly increasing, with isolated zero derivatives, so fixed image values determine convergent labels. Consequently the gradient diverges at fixed $y=(2k+1)\pi +7/6$ as $x\uparrow 1$, and at fixed $y=2k\pi -7/6$ as $x\downarrow -1$, using $k\ge 0$ and $k\ge 1$, respectively, to stay in the half-line initial domain. At these first singularities $y_{\eta \eta }=0$ and $y_{\eta \eta \eta }\ne 0$; this is the onset of loss of invertibility, rather than an ordinary nondegenerate fold at the initial singular time.

The equations of the method of characteristics give $(x,y,u)=(s+t^2/2,t,s{e}^t)$. Their parameter map to the plane has Jacobian determinant one, so it is globally invertible and $u=(x-y^2/2)e^y$ is uniquely determined exactly where $1\le x-y^2/2\le 2$. Both positive and negative $y$ occur. Outside this closed parabolic strip, $u=e^{y}F(x-y^2/2)$ allows arbitrary smooth changes of $F$ outside the initial parameter interval. The boundary characteristic curves carry the data from its endpoints.

The method of characteristics from $s\ne 0$ are $(x,y,z)=(s{e}^t,s{e}^{2t},s^2(2+e^{3t})/3)$. Their projections are half-parabolas $y=x^2/s$ with $x/s>0$. The inverse parameters $s=x^2/y$, $t=\log (y/x)$ exist exactly on the first and third quadrants, giving the unique solution $z=2x^4/(3y^2)+xy/3$ there. The projection Jacobian $s{e}^{3t}$ is nonzero, so every allowed initial point is noncharacteristic.

A smooth characteristic surface in three-dimensional $(x,y,u)$ space need not remain the graph of a differentiable function after projection to the $(x,y)$ plane. Locally, if its implicit equation is $F(x,y,u)=0$, the implicit function theorem produces a smooth graph wherever $F_u\ne 0$. At a point where $F_u=0$, the projection can fold and two graph branches can meet. A real radical formula may continue to that boundary even though its partial derivatives diverge there. Thus the closed set on which the radical is real can be larger than the open domain of a classical solution.

For example,has two branches $u=(15x+12y\pm 5\sqrt{9x^2+40xy})/16$. On them $F_u=\pm 10\sqrt{9x^2+40xy}$, so the graph projection folds on the nonzero boundary points of $9x^2+40xy\ge 0$. Its gradient in the base variables does not vanish at those points, and differentiation of the radical gives the transverse derivative singularity. The characteristic surface itself can be smooth there: its graph coordinate is the part which fails. Boundary values may still be defined continuously, but they do not create an open classical-solution neighborhood.

For $H_t+H^{n+1}{H}_x=1-2x$ with $H(\xi ,0)=0$, method of characteristics conserves $H^{n+2}/(n+2)-(x-x^2)$. A positive downstream branch therefore requires $0\le \xi <x<1-\xi$. Initial points in the negative-accumulation region cannot create nonnegative ice from a bare bed. Characteristic intersections and missing dry-bed physics prevent using the formal initial/boundary branches as an unrestricted global solution.

For a first-order nonlinear partial differential equation $F(x,y,u,p,q)=0$, with $p=u_x,q=u_y$, Charpit's equations extend the method of characteristics to the five-dimensional space of position, value and gradient:The chain rule gives $dF/dt=0$ by direct cancellation, so a trajectory starting on $F=0$ remains there. Initial data $u_0(s)$ on $(x_0(s),y_0(s))$ require both $F(x_0,y_0,u_0,p_0,q_0)=0$ and $p_0{x}_0^{\prime }+q_0{y}_0^{\prime }=u_0^{\prime }$. The non-characteristic condition is $x_0^{\prime }{F}_q-y_0^{\prime }{F}_p\ne 0$. It makes the projection onto $(x,y)$ locally invertible. Compatible initial contact data then propagate as $du=pdx+qdy$, reconstructing a classical solution. Multiple permitted initial normal derivatives can give different local branches; loss of the projection Jacobian determinant marks a characteristic caustic.

For initial data $u(x,0)=1+x^3/4$ on $0\le x\le 1$, the characteristic surface has $x=s{e}^r$, $y=3s^2(1-e^{-2r})/2$, $u=1+s^3(e^{-3r}-3e^{-r}/4)$. Its projection determinant is $3s^2{e}^r(2e^{-2r}-1)$, so the sheet connected to the initial data loses local invertibility at $r=\log 2/2$. The regular sheet has $r<\log 2/2$ and projects to $0<x<\sqrt{2}$, $3(1-x^{-2})/2<y<3x^2/8$. Its upper boundary is a fold point of a critical manifold caustic. Solving the inverse yields $s^2=[x^2+\sqrt{x^4-(8/3)x^{2}y}]/2$; the other square-root branch lies on the folded sheet and cannot simply be joined as a unique smooth solution. The initial point $s=0$ is already characteristic, illustrating the need to inspect endpoints as well as the interior of a Cauchy curve.

For a fixed real unit vector $k$, the linear ordinary differential equation $\dot{r}=r-k\times r$ combines dilation with rotation about the axis $\mathbb{R}k$. Its solution from a line $s\ell$, with $|\ell |=1$, isSince $k\cdot r=s{e}^t\kappa$ and $|r|=|s|e^t$, the union of these characteristic curves is given by $|k\cdot r|=|\kappa ||r|$. For $0<|\kappa |<1$ it is a circular cone of semi-angle $\arccos |\kappa |$: varying $t$ supplies every azimuth and varying $s$ supplies every radius on both halves. For $\kappa =0$ it is a plane; for $|\kappa |=1$ it is only the axis. In the last case the line is itself a characteristic curve, so the initial data do not generate a two-dimensional surface. The cone is a smooth surface away from its vertex, and it is a graph of a function only on suitable patches. Invariance of a graph patch under characteristic curves holds only while the curve stays in its domain; global flow invariance requires continuation of that patch.

A characteristic caustic is a locus where the coordinate map from initial-data and evolution parameters to physical coordinates has zero Jacobian determinant. The inverse function theorem then fails, and different characteristic branches can reach the same point. A classical solution constructed from initial data must be restricted to its appropriate invertible branch before such a fold. In geometrical optics the corresponding ray-envelope phenomenon is a caustic; in nonlinear first-order equations a fold can signal the onset of multiple gradients or shock formation by characteristic intersection.

For a first-order partial differential equation with the displayed projected characteristic field, label the initial segment by $r=y(0)\in [0,c]$. The characteristic curves are $x=a-(a-b)e^s$, $y=r{e}^s$. They are the branches of straight lines through $(a,0)$ on the side $x<a$. The inverse coordinates areThey are unique exactly in the displayed sector. The Jacobian determinant is $-(a-b)e^{2s}\ne 0$, so the initial segment is noncharacteristic and its transported data determine a unique classical solution on the sector's interior whenever the characteristic equation for the dependent variable has a unique solution. The bounding rays inherit data from the two endpoints. The point $(a,0)$ is approached only at infinite parameter time, and characteristics on the opposite side cannot reach the initial segment.

For the first-order quasilinear partial differential equation $P(x,y,z)z_x+Q(x,y,z)z_y=R(x,y,z)$, the displayed ordinary differential equations trace curves tangent to every classical solution surface. Indeed, if $z=u(x,y)$ solves the equation and $x_s=P$, $y_s=Q$, then the chain rule gives $z_s=u_{x}P+u_{y}Q=R$. Conversely a surface swept out by these curves solves the equation wherever its projection to $(x,y)$ is invertible and hence represents a differentiable graph. The coefficients can depend on the unknown $z$, which distinguishes these equations from characteristic equations for a prescribed linear transport equation.

A characteristic surface may remain smooth while its projection to the independent-variable plane fails to be locally invertible. At a generic fold, smooth changes of parameter and plane coordinates reduce this projection to $(u,v)\mapsto (u,v^2)$. There are two sheets on one side of the image of $v=0$, no sheet on the other, and one merged sheet on the fold itself. The image is a fold caustic. A characteristic graph can terminate there even though the space curves defining the surface continue smoothly. Vanishing of the Jacobian determinant detects a candidate singular locus; verifying the fold requires a nonzero quadratic term in its null direction, as in $v^2$.

For the linear partial differential equation $2y{u}_x-x{u}_y=0$, the method of characteristics gives $\dot{x}=2y$, $\dot{y}=-x$, $\dot{u}=0$. Thus $\varphi =x^2+2y^2$ is constant on each characteristic curve, whose nonzero level sets are ellipses. On a domain containing entire such ellipses, the solutions have the form $u=F(\varphi )$, subject to the desired regularity at the origin.

For the inhomogeneous equation $2y{u}_x-x{u}_y=xy(2y^2-x^2)$, a particular solution is $u_p=x^2{y}^2/2$, so the general solution is $u=x^2{y}^2/2+F(x^2+2y^2)$. Prescribing $u(x,1)=x^2$ fixes $F(s)=(s-2)/2$ only for $s\ge 2$. The inner ellipses $s<2$ never intersect that data line, so the boundary values leave their solution values undetermined. Smooth extensions must match the prescribed $F$ at $s=2$; choosing the same linear formula everywhere gives one polynomial extension.

The characteristic ordinary differential equations $\dot{x}=y$, $\dot{y}=x$ preserve $u=x^2-y^2$. Away from the origin, $u=x^2-y^2$, $v=2xy$ have nonzero Jacobian determinant $4(x^2+y^2)$, and the chain rule reduces the displayed equation to $2(x^2+y^2)f_v=0$. Hence locally $f=F(u)$.

Globally, a level set of $u$ can have disconnected characteristic components, permitting different profiles on its different branches. A smooth counterexample on the whole plane is $f=\exp [-1/(x^2-y^2{)}^2]$ where $x>0$ and $x^2>y^2$, and $f=0$ elsewhere. The exponential is flat on the wedge boundary, so this extension is smooth and satisfies the equation. Nevertheless the points $(1,0)$ and $(-1,0)$ have equal $u$ and different function values. One global profile $F(u)$ therefore needs an additional branch-identification condition, rather than following from the local change of variables alone.

For $x{f}_x+\beta y{f}_y=\gamma f$, the curves in the method of characteristics are $x=x_0{e}^s$, $y=y_0{e}^{\beta s}$ and $f=f_0{e}^{\gamma s}$. On either component $x\ne 0$, the invariant is $y|x{|}^{-\beta }$, giving $f=|x{|}^{\gamma }F(y|x{|}^{-\beta })$. A coordinate change that includes $y/x$ may be singular on an axis, so smooth continuation there must be checked against the original partial differential equation.

If initial data are parameterized by $s$ and their characteristic invariant is $D(s)$, a solution constant on characteristics requires $f(s)=F(D(s))$. At a fold $D^{\prime }(s_0)=0$, equal invariant values must have equal data, and $f^{\prime }(s)/D^{\prime }(s)$ must extend continuously if $F$ is continuously differentiable. The condition $f^{\prime }(s_0)=0$ alone is insufficient: for $D(s)=1/4-(s-1/2{)}^2$, the data $|s-1/2{|}^{3/2}$ would require a derivative of $F$ blowing up at $1/4$.

For a transport equation ${\partial }_{t}f+b(t,z)\cdot {\nabla }_{z}f=h$, solve $\dot{z}=b(t,z)$. The chain rule then gives $d[f(t,z(t))]/dt=h(t,z(t))$. Initial data and the time integral of this source determine the solution along each characteristic curve.

A linear transport equation has a prescribed velocity field and is linear in the transported unknown.

For $a_t+(V(s)a{)}_s=0$ with $V>0$, the flux perturbation $b=Va$ solves $b_t+V{b}_s=0$. The travel coordinate transforms it into unit-speed transport. Thus the flux perturbation is constant on characteristics while area amplitude changes as $1/V$. In a supplied power-law river, $A_{\mathrm{eq}}^{\prime }=M/V$, giving travel time $[A_{\mathrm{eq}}(l)-A_{\mathrm{eq}}(0)]/M$. At a zero-speed head, this time can remain integrable even though the amplitude formula is not uniform.

For the constant-velocity linear transport equation $u_t+c\cdot {\nabla }_{x}u=0$, a Dirac delta distribution initially at zero moves along $x=ct$. Its spacetime distribution pairs by $⟨u,\phi ⟩=\int \phi (ct,t)dt$, and its initial trace is obtained by weak convergence of distributions. The Fourier representation of the Dirac delta function writes it as an oscillatory integral with phase function $(x-ct)\cdot \theta$. Its singular support is precisely the trajectory; it has order-zero distribution regularity as a measure, without becoming a smooth function.

If a velocity operator satisfies $\mathrm{Re}⟨Lh,h⟩\ge \delta \Vert h{\Vert }_2^2$, then sufficiently regular solutions of ${\partial }_{t}f+v\cdot {\nabla }_{x}f+Lf=0$ obey the displayed estimate when spatial boundary fluxes vanish. The free-transport part contributes zero to the energy derivative. Integrating the resulting differential inequality avoids dividing by a possibly zero norm.

Scalar advection preserves every finite Lp norm if its complete differentiable characteristic flow preserves volume and the initial datum belongs to the corresponding Lp space. In general the $p$th power of the norm is the initial $|f_0{|}^p$ weighted by the phase-space flow Jacobian. Completeness alone preserves the essential supremum, but does not preserve finite-$p$ norms for compressible transport.

For a transport equation on a domain, prescribe boundary data only where the velocity points into the domain. On $x>0$ with speed $a>0$, a bounded weak solution with initial data $u_0$ and inflow data $f$ satisfiesThe integrals are over the quadrant and its two boundary rays. Backtracking characteristic curves gives $u_0(x-at)$ when $x\ge at$ and $f(t-x/a)$ otherwise. Corner values need not match for a bounded weak solution.

For the transport equation $u_t+a{u}_x=0$ on the quadrant with $a>0$, smooth initial and inflow data glue across $x=at$ to order $m$ precisely when $f^{(j)}(0)=(-a{)}^j{u}_0^{(j)}(0)$ for $0\le j\le m$. Matching all orders yields a smooth solution; matching orders zero and one yields a classical solution up to the two boundaries.

For a complete $C^1$ velocity flow, the bounded weak solution of $u_t+F\cdot \nabla u=0$ is its initial value pulled back by the inverse flow. Its weak identity is $\int u({\phi }_t+\mathrm{div}(F\phi ))+\int u_0\phi (0)=0$; the divergence term is essential. Changing to flow coordinates gives $\int w{\partial }_t(J\zeta )+\int u_0\zeta (0)=0$. Approximation in the spatial parameter permits $\zeta =\eta /J$, since $J$ and $J_t$ are continuous and $J>0$ on compact sets. Hence ${\partial }_{t}w=0$ distributionally, proving uniqueness without requiring second derivatives of $F$.

An adjoint transport equation transfers the transport operator from a weak solution to a test function. Solving the terminal-value adjoint equation allows arbitrary compactly supported source terms to be used in uniqueness proofs.

A system of conservation laws evolves a vector of conserved variables $U$ through a flux vector $F(U)$. For a smooth solution it has quasilinear form $U_t+A(U)U_x=0$, where $A(U)=DF(U)$ is the flux Jacobian.

Two smooth differential systems related by a nonlinear change of dependent variables need not define the same weak solutions. If $U_t+F(U{)}_x=0$, a smooth function $V=g(U)$ satisfies $V_t+g^{\prime }(U)F^{\prime }(U)U_x=0$ for smooth $U$; even when this can be written as another divergence, its Rankine-Hugoniot conditions need not be equivalent to those in $U$. The smooth chain rule does not justify multiplying jump distributions by discontinuous coefficients.

For example the shallow water equations can be written smoothly as $h_t+(hv{)}_x=0$ and $v_t+(h+v^2/2{)}_x=0$. Mechanical mass and momentum conservation instead select $h_t+(hv{)}_x=0$ and $(hv{)}_t+(h{v}^2+h^2/2{)}_x=0$. These systems agree for smooth positive $h$, but their jump laws differ. The states $(h,v)=(2,\sqrt{3}/2)$ and $(1,0)$ satisfy the mass/momentum jump laws with speed $\sqrt{3}$, while the velocity-form jump equation would demand $11/(4\sqrt{3})$. A model for discontinuities must specify the conservative variables or an equivalent admissible regularization, rather than using smooth equivalence alone.

For a strictly hyperbolic system of conservation laws $U_t+F(U{)}_x=0$ with ordered characteristic speeds ${\lambda }_1<\cdots <{\lambda }_n$, a Lax family-$j$ shock has ${\lambda }_j(U_R)<s<{\lambda }_j(U_L)$. The remaining families must stay on the same side of $s$: ${\lambda }_{j-1}(U_L)<s$ and $s<{\lambda }_{j+1}(U_R)$ where those neighbors exist. These inequalities imply that the lower families have speed below $s$ on both sides and the higher families speed above $s$ on both sides. There are $n+1$ incoming characteristic directions: two from the compressed family and one from each other family. The Rankine-Hugoniot conditions supply the jump balance, while these inequalities supply admissibility. For a convex scalar flux this reduces to the entropy shock condition.

A Riemann problem for a one-dimensional system of conservation laws ${\partial }_{t}U+{\partial }_{x}f(U)=0$ has two constant initial states, $U_L$ for $x<0$ and $U_R$ for $x>0$. The data and equation are invariant under $(x,t)\mapsto (\lambda x,\lambda t)$. When the physically admissible entropy solution is unique, this invariance makes it a self-similar solution depending on $x/t$. For ideal-gas compressible flow, the characteristic speeds are $u-c,u,u+c$. The two acoustic families carry shock waves or rarefaction waves, while the middle family carries a contact discontinuity. Sufficiently strong separating velocities can instead create a vacuum region.

Two positive uniform ideal-gas states can be joined by one nontrivial rarefaction wave precisely when their specific entropy agrees, $p_L/{\rho }_L^{\gamma }=p_R/{\rho }_R^{\gamma }$, and for one $\sigma \in \{\pm 1\}$ they satisfy $u_R-u_L=2\sigma (c_R-c_L)/(\gamma -1)$ with $\sigma (c_R-c_L)>0$. These equations are the constant-entropy and Riemann invariant conditions of the self-similar ideal-gas rarefaction. They are sufficient because its affine $u,c$ profiles match the two states at ${\xi }_L=u_L+\sigma c_L$ and ${\xi }_R=u_R+\sigma c_R$, and ${\xi }_R>{\xi }_L$. The left family has $c_R<c_L$, while the right family has $c_R>c_L$. The inequality $u_R>u_L$ alone does not impose the two matching equalities.

A contact discontinuity in one-dimensional gas dynamics moves with the common fluid velocity $s=u_L=u_R$ and has continuous pressure, while mass density and specific entropy can jump. The mass jump condition $s[\rho ]=[\rho u]$ is automatic with common $u=s$; momentum conservation $s[\rho u]=[\rho u^2+p]$ then requires $[p]=0$. The energy jump condition is also satisfied. Thus a contact is a material interface, not a compression shock. A smooth self-similar positive-density flow cannot spread this zero-speed-relative-to-the-fluid family into a fan: $u=x/t$ on an interval would require both $du/d(x/t)=1$ and, from continuity, $du/d(x/t)=0$.