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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.

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.

The total differential order of ${\partial }_t-{\partial }_x^2$ is two, so its principal symbol is $p(\tau ,\xi )=-{\xi }^2$. The conormal to $t=0$ is $(1,0)$, on which $p$ vanishes. The initial line is a characteristic hypersurface for this total-order symbol; the first-order time derivative does not enter it.

Suppose a real analytic solution existed near $(0,0)$. Repeated use of the heat equation gives ${\partial }_t^{k}u={\partial }_x^{2k}u$. The initial power series is ${\sum }_{j\ge 0}(-1{)}^j{x}^{2j}$ near zero, henceThe time Taylor series at $x=0$ would therefore have coefficients $(-1{)}^k(2k)!/k!$. The ratio of successive absolute coefficients is $2(2k+1)\to \infty$, giving radius of convergence zero. This contradicts the assumed real analytic regularity. No such analytic local solution exists, although the initial function itself is real analytic.

For a defining function $\varphi$ with $d\varphi \ne 0$, the characteristic hypersurface test is that the principal symbol vanish at $d\varphi$.

For the wave equation with speed $c>0$,Thus its characteristic hypersurfaces satisfy ${\varphi }_t^2=c^2|{\nabla }_x\varphi {|}^2$. In one space dimension the two families are $x\pm ct=\text{constant}$; cones are characteristic away from their vertices.

For the free Schrodinger equation, in normalized units,Its total-order characteristic hypersurfaces satisfy ${\nabla }_x\varphi =0$. Their normal is purely temporal, so locally they are constant-time hypersurfaces. Multiplying the equation by a nonzero constant or choosing the opposite sign convention does not change this test.

For the Laplace equation,There are no real characteristic hypersurfaces for the Laplace equation, since their normal cannot be zero. These statements concern the ordinary total-order principal symbol, not a weighted space-time grading.

For the heat equation, viewed as a second-order equation in all $(t,x)$ variables, the principal symbol is $|\xi {|}^2$. It vanishes on $dt$, so $t=0$ is a characteristic hypersurface andWriting $u_t={\Delta }_{x}u$ does not repair its hypotheses: the right side has second spatial derivatives, exceeding the first-order normal-form allowance. Indeed real analytic initial data need not produce a jointly real analytic solution through time zero. In one dimension, $u_0(x)=(1+x^2{)}^{-1}$ would forceThe time Taylor coefficients $(-1{)}^k(2k)!/k!$ have zero radius of convergence.

Nevertheless the forward heat equation has Hadamard well-posedness in standard $L^2$ or Sobolev space classes for $t\ge 0$; its Fourier multiplier $e^{-t|\xi {|}^2}$ gives stable smoothing. Backward evolution has the growing multiplier $e^{|t||\xi {|}^2}$ and is unstable. Neither forward stability nor backward instability is decided simply by whether the real analytic theorem applies.

For $F(x,t)=t-\tau (x)$, the conormal to its zero set is $dF=(-D\tau ,1)$. The principal symbol of ${\partial }_t^2+L$ isA characteristic hypersurface requires $p(x,t;dF)=0$ on the surface. Thus $\tau$ satisfies the anisotropic eikonal equationThe first-order and zeroth-order terms do not affect this condition.

Expanding the equation givesFor the hyperbolic partial differential equation discriminant, $A=1$, $B=\rho$, $C={\rho }^2-1$, soFor a curve $\rho =\rho (t)$, the characteristic hypersurface equation is $({\rho }^{\prime }{)}^2-2\rho {\rho }^{\prime }+{\rho }^2-1=0$, or ${\rho }^{\prime }=\rho \pm 1$. Therefore the two families of characteristic curves areThe values $C=0$ include the characteristic lines $\rho =-1$ and $\rho =1$. Every curve in the first family approaches $\rho =-1$ as $t\to -\infty$; every curve in the second approaches $\rho =1$. The two curves through $(\rho ,t)=(0,0)$ are $\rho =e^t-1$ and $\rho =1-e^t$, shown as the thick curves below. Equivalently, the characteristic coordinates are $e^{-t}(\rho +1)$ and $e^{-t}(\rho -1)$.

Locally write the real analytic hypersurface as $\Sigma =\{F=0\}$ with $dF\ne 0$. The conormal bundle is spanned by $dF$. The surface is a characteristic hypersurface at $p$ precisely when the principal symbol vanishes on that conormal:Only the matrix $(a_{ij}+a_{ji})/2$, which is a symmetric matrix, contributes to this expression.

For $\varphi =t-x_1$, the conormal is $dt-d{x}_1$, andThis null hyperplane is a characteristic hypersurface of the wave equation.

A characteristic hypersurface of the form $\varphi (t,x)=p(t)\pm q(x)$ must satisfyOn either connected component of $tx>0$, separation givesOne choice valid on both components isIndeed, $p^{\prime }(t)=|t{|}^{-1/2}$ and $q^{\prime }(x)=|x{|}^{1/2}$ away from the axes, so $tx{p}^{\prime 2}=q^{\prime 2}$. Thusare the two families of characteristics, and $p\pm q$ are characteristic coordinates.