Characteristic hypersurface

A hypersurface $\Sigma =\{\varphi =0\}$ is characteristic for a differential equation at $x\in \Sigma$ when its principal symbol vanishes on the conormal $d\varphi (x)$:It is non-characteristic where this quantity is nonzero.

For a scalar second-order partial differential equation in two variables, a tangent direction along which Cauchy data fail to determine the highest transverse derivative through the equation. For principal part $A{u}_{xx}+2B{u}_{xy}+C{u}_{yy}$, a tangent $(dx,dy)$ is characteristic exactly when $A(dy{)}^2-2Bdxdy+C(dx{)}^2=0$. Equivalently its conormal annihilates the principal symbol. A hyperbolic partial differential equation has two distinct real characteristic directions at each point.

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.

A characteristic plane is a planar characteristic hypersurface. For $u_{zz}-m^2{u}_{xx}=0$, the planes $x\pm mz=\text{constant}$ carry the two travelling-wave families.

A non-characteristic hypersurface satisfies $p(x,d\varphi (x))\ne 0$. This lets the equation solve for the highest derivative normal to the hypersurface.

A characteristic coordinate is a function constant along one family of characteristics. For a second-order equation in two variables, two independent characteristic coordinates reduce the principal part to a mixed derivative.

For a hyperbolic partial differential equation in two variables, two independent characteristic coordinates remove both pure second partial derivatives, leaving a nonzero mixed second partial derivative. A nonzero Jacobian determinant is essential: two characteristic invariants that become dependent do not give a valid coordinate transformation.

In $x,y>0$, the change of variables in the title reducesto the one-dimensional wave equation $U_{ss}-U_{rr}=0$. Indeed $u_{xx}=U_{rr}/(4x)-U_r/(4x^{3/2})$ and $u_{yy}=y{U}_{ss}/4+U_s/(4\sqrt{y})$; the first-derivative terms cancel exactly against the right-hand side. The independent characteristic coordinates $\xi =r+s$, $\eta =r-s$ then give $V_{\xi \eta }=0$ and $u=g(\xi )+h(\eta )$. Two characteristic families alone generally reduce only the principal part; the cancellation of lower-order terms is the extra feature that gives this simple exact solution.

The change of variables $X=\rho e^{-t}$, $S=e^{-t}>0$ turnsinto $w_{SS}-w_{XX}=0$. Indeed, with $D={\partial }_t+\rho {\partial }_{\rho }$, the original operator is $D^2+D-{\partial }_{\rho }^2$, while $D=-S{\partial }_S$ and ${\partial }_{\rho }=S{\partial }_X$. Its characteristic curves are $\rho =-1+C{e}^t$ and $\rho =1+C{e}^t$, including the characteristic lines $\rho =\pm 1$. General smooth solutions have the form $F(e^{-t}(\rho +1))+G(e^{-t}(\rho -1))$. Zero Cauchy data on $-1<\rho <1$ at $t=0$ force vanishing throughout that strip for $t\ge 0$, but not for all negative time: $u=(e^{-t}(\rho +1)-2{)}_{+}^3$ is a global $C^2$ counterexample. Thus the characteristic barriers describe a forward domain of dependence, not a barrier in both time directions.