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

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 $x{f}_x+\beta y{f}_y=\gamma f$, 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.