Characteristic curve

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 converts a first-order partial differential equation into ordinary differential equations along characteristic curves. Forsolve $\dot{X}=b(t,X)$, $X(0)=\alpha$, and $\dot{U}=h(t,X,U)$, $U(0)=g(\alpha )$. The chain rule identifies $U(t,\alpha )=u(t,X(t,\alpha ))$. If the characteristic flow map is invertible, it reconstructs the solution by $u(t,x)=U(t,X_t^{-1}(x))$. This turns spatial propagation and the evolution of the carried value into separate calculations.^[ref]

For $u_t+t{u}_x=0$, the characteristic curves are $X(t,\alpha )=\alpha +t^2/2$ and $U=g(\alpha )$, so $u(t,x)=g(x-t^2/2)$. Substitution verifies both the partial differential equation and the initial condition. More generally, a quasilinear partial differential equation allows $b$ to depend on $U$, so the two ordinary differential equations become coupled. Noncharacteristic Cauchy data and a nonzero projection Jacobian determinant give local reconstruction; characteristic crossing can destroy it even while the lifted trajectories remain smooth.

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.