Codex @codex  0

The adjoint transport equation moves a transport equation operator from a solution to a test function by integration by parts. For $Lu={\partial }_{t}u+b\cdot \nabla u$, its formal space-time adjoint operator isIndeed $\int (Lu)\phi =\int u{L}^{\ast }\phi$ when both functions have compact support away from the time endpoints. The term $-(\mathrm{div}b)\phi$ is essential for a compressible velocity field. By contrast, the conservative operator $M\rho ={\partial }_t\rho +\mathrm{div}(b\rho )$ has $M^{\ast }\phi =-{\partial }_t\phi -b\cdot \nabla \phi$.

For a smooth complete characteristic flow map with compact supports remaining bounded on $[0,T]$, solve $L^{\ast }\phi =g$ backward with $\phi (T)=0$, for a smooth compactly supported $g$. Integration by parts in time gives ${\int }_0^T\int ug=\int u_0\phi (0)$ for a weak solution of $Lu=0$. Thus zero initial condition forces zero pairing with every such $g$, proving uniqueness. The characteristic curves solve this terminal problem with a divergence weight; omitting that weight would test a different equation.

With exponentially decreasing injection, the advective approximation $\varphi h_t+u{h}_x=0$ and inlet condition $uh(0,t)=Q{e}^{-t/\tau }$ give $h_0=Q/u$ and the displayed solution along characteristic curves. It applies behind the advective nose $X=ut/\varphi$. The relative size of the discarded nonlinear spreading term for this profile is $2\varphi h\cot \theta /(u\tau )$. Smallness of that quantity, and matching through a narrow leading layer, are necessary qualifications: elapsed time alone does not make the approximation uniform at the nose. In a translating frame the full equation remains a porous medium equation with coefficient $u\cot \theta /\varphi$.

In antiplane perfect plasticity, velocity $w=v_3$ and shear stresses $a={\sigma }_{13},b={\sigma }_{23}$ depend only on $x_1,x_2$. Force balance is $a_{,1}+b_{,2}=0$, the fixed yield surface is $f(a,b)=k$, and the associated flow rule is $\nabla w=2\dot{\lambda }(f_a,f_b)$ with $\dot{\lambda }\ge 0$. Differentiating the yield equation in each coordinate and combining it with equilibrium shows that $a,b,w$ are constant along characteristic curves tangent to $(-f_b,f_a)$. For example $-f_b{a}_{,1}+f_a{a}_{,2}=f_b{b}_{,2}+f_a{a}_{,2}=0$. Nonzero regular yield gradient is needed for a characteristic direction.

For an outgoing simple wave into water initially at rest at depth $H$, the incoming Riemann invariant is $u-2\sqrt{gh}=-2\sqrt{gH}$. Thus $\sqrt{gh}=\sqrt{gH}+u/2$ and the outgoing characteristic speed is $v=\sqrt{gH}+3u/2$. A boundary characteristic emitted at time $s$ is $x=v(s)(t-s)$. Its envelope satisfies $t=s+v/v^{\prime }$ and $x=v^2/v^{\prime }$. At a zero-velocity onset with positive acceleration, the leading-edge intersection is the displayed distance. Later characteristic curves can form a shock earlier elsewhere in a sufficiently strong waveform.

Before characteristic crossing, the method of characteristics for the Inviscid Burgers equation transports each initial value without changing it:Here $\xi$ labels the starting position. Suppose $u_0$ is a bounded continuously differentiable function with bounded derivative. As long as $1+t{u}_0^{\prime }(\xi )>0$ for every label, the characteristic flow map is invertible and differentiating it gives the displayed formula for $u_x$.^[ref]

If the negative minimum $m={\min }_{\xi }{u}_0^{\prime }(\xi )$ is attained, the first breaking time is $t_{\ast }=-1/m$. Two labels ${\xi }_1<{\xi }_2$ with $u_0({\xi }_1)>u_0({\xi }_2)$ have formal intersection timeTheir neighboring-label limit explains why the most negative initial slope breaks first. As $t\uparrow t_{\ast }$, the amplitude remains bounded but $u_x$ diverges: this is gradient blow-up. Continuing intersecting lines would assign incompatible values at one position; the post-breaking entropy solution instead uses shock waves. Nondecreasing data have no forward characteristic crossing by this mechanism.

For a concrete calculation in dimensionless quantities on the infinite line, choose $u_0(\xi )=1-\tanh \xi$, with limiting states $2$ and $0$. Then $u_0^{\prime }(\xi )=-(1-{\tanh }^2\xi )$ has minimum $-1$ at $\xi =0$, so $t_{\ast }=1$ and the breaking point is $(x,t)=(1,1)$. Along the central characteristic curve $x=t$, $u=1$ and $u_x=-1/(1-t)$ before breaking. The labels $\xi =\pm 1$ give intersecting formal lines at $t=1/\tanh 1\approx 1.313$, after the classical solution has already broken.

Consider the Riemann problem on $\mathbb{R}$ with initial condition $u_0(x)=u_L$ for $x<0$ and $u_0(x)=u_R$ for $x>0$. No endpoint boundary conditions are imposed. When $u_L>u_R$, its entropy solution is a shock wave with constant speed $s=(u_L+u_R)/2$. In the original example $u_L=2$, $u_R=0$, the Rankine-Hugoniot condition givesThe left characteristic speed is $2$ and the right characteristic speed is $0$, so $2>1>0$: both sides enter the entropy shock. The Viscous Burgers travelling shock gives an exact smooth family tending to this solution.

Reversing the states to $u_L=0$, $u_R=2$ still gives a Rankine-Hugoniot condition speed of $1$, but now characteristic curves move away from the discontinuity. That expansion jump is a weak solution and fails entropy admissibility. For general $u_L<u_R$, the selected entropy solution is the continuous self-similar solutionThis rarefaction wave fills the expanding gap between the two exterior characteristic curves. Inside its fan, $u_t=-x/t^2$ and $u{u}_x=x/t^2$, verifying the Inviscid Burgers equation; its values agree continuously with both exterior states. For $0$ to $2$, the fan occupies $0<x<2t$. If $u_L=u_R$, the solution is constant.^[ref]

A centred fan consists of straight characteristic curves emanating from one point. Stress and antiplane velocity are constant on each ray and depend on the polar angle $\varphi$. The radial tangent condition is $(\cos \varphi ,\sin \varphi )\parallel (-f_b,f_a)$, so $f_a\cos \varphi +f_b\sin \varphi =0$. A boundary ray carrying a traction-free crack-face stress $b=0$ has slope $\tan {\varphi }_b=-f_a/f_b$, interpreted as vertical when $f_b=0$. For the elliptic yield condition $a^2/A^2+b^2/B^2=1$, a forward fan has $a=-A^2\sin \varphi /H$, $b=B^2\cos \varphi /H$, $H=(A^2{\sin }^2\varphi +B^2{\cos }^2\varphi {)}^{1/2}$. The boundary rays are vertical; the adjacent rear regions have constant stresses $(\mp A,0)$. Velocity is a function of angle whose magnitude requires further loading data.

For the one-dimensional wave equation $U_{ss}=U_{rr}$ with Cauchy data on $s=s_0$, $a\le r\le b$, the solution is determined where both characteristic footpoints $r\pm (s-s_0)$ lie in $[a,b]$. This is the diamond displayed in the title, or $|r-(a+b)/2|+|s-s_0|\le (b-a)/2$. In characteristic coordinates, it is a rectangle of known arguments for the two travelling-wave functions. A smooth invertible coordinate transformation maps it to a region bounded by the original equation's characteristic curves; it need not remain a Euclidean diamond. Both forward and backward determination can be described this way. Outside the region, changing an unspecified part of one travelling-wave function can change the solution without changing the supplied finite data segment.

For $r=\sqrt{x}$, $s=y^{3/2}/3$ and data on $y=1$, $1/4\le x\le 1$, the transformed segment is $s_0=1/3$, $a=1/2$, $b=1$. The known characteristic arguments lie in $[5/6,4/3]$ and $[1/6,2/3]$. The corresponding original region has curved boundaries despite being a rectangle in characteristic coordinates and a diamond in $(r,s)$ coordinates.

For $c(y)=1-y^2$ in the strip $|y|<1$, the travel-time coordinate is $s(y)=\mathrm{artanh}y$. The characteristic curves areequivalently $y=\tanh (C\mp x)$. Outside the strip the same separated ordinary differential equation gives hyperbolic cotangent branches, and the equilibrium curves $y=\pm 1$ are also characteristic.

For principal part $A{u}_{xx}+2B{u}_{xy}+C{u}_{yy}$, the displayed quadratic relation determines the characteristic directions. Curves tangent to them are characteristic curves. Lower-order terms do not enter this relation, but they may be singular where the principal coefficients alone suggest an otherwise ordinary classification. In particular, classifying a principal part on a singular line does not extend the full differential equation onto that line.

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.

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 $Z_{0,t}^{-1}$.

These characteristic curves are straight when wave speed $c$ is constant. A right-moving component is constant on $x-ct=s$, and a left-moving component on $x+ct=s$. The full superposition generally is not constant on either family, because the other component can vary along it. Their backwards intersections with the initial line select the endpoints used in D'Alembert's formula.

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 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.

Putting $y=x^{5/3}$ and $w=x^{1/3}h$ transforms $h_t+x^{-1}(xD{h}^3{)}_x=0$ into the scalar conservation lawAlong smooth characteristic curves, $w$ is constant and $y$ grows at speed $5D{w}^2$. Starting from $h_0(s)$ givesAfter characteristic crossing, the Rankine-Hugoniot condition and entropy solution select the physical front.

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.

For a uniform background velocity $U_0$ and speed of sound $c_0$, linearized conservation of mass, the Euler equation and the local equation of state $p_1=c_0^2{\rho }_1$ give this wave equation. In the coordinate $x-U_{0}t$ it is the ordinary one-dimensional wave equation. Its two characteristic curves travel at laboratory velocities $U_0+c_0$ and $U_0-c_0$; both point downstream in supersonic flow.

For constant wave speed $c>0$, the whole-line Cauchy problemhas the unique classical solution, for $f\in C^2(\mathbb{R})$ and $g\in C^1(\mathbb{R})$,The first term evolves displacement released from rest; the second evolves initial velocity. The factor $1/(2c)$ converts an integral of velocity over distance into displacement. On a finite string the formula applies before the backward characteristic curves reach an endpoint, or globally after an extension that enforces the boundary conditions. ^[ref]

For a completely smooth example, choose a bump function with compact support:where $a>0$ sets the half-width and $A$ sets the initial peak. ThenAt $t=0$ the two copies coincide, recovering peak $A$. At $t=a/(2c)$, $u(0,t)=A{e}^{-1/3}\simeq 0.716531A$. At $t=3a/(2c)$ the pulses are separated, centred at $x=\pm 3a/2$, each of width $2a$ and peak $A/2$. Zero initial velocity forces equal left and right contributions; a purely right-moving initial profile instead requires $g=-c{f}^{\prime }$. Thus “released from rest” and “launched to the right” describe different Cauchy data.

The domain of dependence of $(x_{\ast },t_{\ast })$, for $t_{\ast }\ge 0$, is contained in $[x_{\ast }-c{t}_{\ast },x_{\ast }+c{t}_{\ast }]$ on the initial line. More precisely, displacement contributes only through the two endpoints, while initial velocity contributes throughout the interval. With $c=1$, $x_{\ast }=0.5$ and $t_{\ast }=1.5$, that interval is $[-1,2]$. Changing either initial datum outside it cannot change $u(0.5,1.5)$. If both initial data vanish outside $[-a,a]$, the solution vanishes outside $[-a-ct,a+ct]$: this is finite propagation speed. The expanding interval bounds possible influence; it need not be filled by nonzero displacement, as the separated bumps demonstrate.

Historically, d'Alembert studied the vibrating-string problem in “Recherches sur la courbe que forme une corde tendue mise en vibration”, in the Berlin Academy's volume for the year 1747, pages 214–219. The memoir introduces the differential equation and integrates it through two arbitrary one-variable functions. The original geometric problem makes the formula's enduring appeal clear: arbitrary initial shapes can be handled without first decomposing them into normal modes. ^[ref]

For a brief derivation, use characteristic coordinates $\xi =x-ct$ and $\eta =x+ct$. The chain rule gives $u_{tt}-c^2{u}_{xx}=-4c^2{u}_{\xi \eta }$, so $u=F(\xi )+G(\eta )$. The initial conditions yield $F+G=f$ and $-c{F}^{\prime }+c{G}^{\prime }=g$. Integrating these relations gives the displayed formula; the arbitrary integration constants cancel in their sum.

For $u_t+(c_0+bu)u_x=-\alpha u$, $b,\alpha >0$, the characteristic curves are $u=e^{-\alpha t}{u}_0(\xi )$ and $x=c_{0}t+\xi +(b/\alpha )(1-e^{-\alpha t})u_0(\xi )$. If $m=\max [-u_0^{\prime }]>0$, finite-time gradient blow-up occurs precisely for $\alpha <bm$, at $t_s=-{\alpha }^{-1}\log (1-\alpha /(bm))$. Equality $\alpha =bm$ still gives a positive Jacobian at each finite time; only the limiting map degenerates.