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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.

The method of characteristics replaces the first-order partial differential equation by the ordinary differential equationsAn integrating factor for the second equation gives $d(e^{-2s}y)/ds=\beta C{e}^{-s}$, so their general solution of a differential equation isHere $C,D,E$ are real constants. For $C\ne 0$, eliminating $s$ gives the characteristic projectionseach restricted to either $x>0$ or $x<0$. These are arcs of parabolas, with the straight-line exception $K=0$. Along an oriented characteristic curve, $|x|$ increases with $s$.

For $C=0$, the characteristic projections lie on $x=0$: $y=D{e}^{2s}$ describes either vertical half-line when $D\ne 0$. For $C=D=0$ the projection is the single point $(0,0)$, although $u=2s+E$ still changes. This last lifted characteristic curve cannot be contained in the function graph of a classical solution: at $(0,0)$ the original partial differential equation would require $0=2$.

For $\beta =0$, the equation for the characteristic curves isThe two families are therefore $y=\text{constant}$ and the characteristic projections from part (a), $y=K{x}^2$, together with the vertical characteristic $x=0$. On either region $y>0$ or $y<0$, the hyperbolic partial differential equation admits the characteristic coordinatesIndeed $\phi$ is constant along the first family, while $(x{\partial }_x+2y{\partial }_y)\psi =0$. Their Jacobian determinant isThe coefficients of the two pure second partial derivatives in these coordinates vanish, and the coefficient of the mixed partial derivative is $2y/\sqrt{|y|}\ne 0$. Hence they give the hyperbolic characteristic normal form. They also remain regular at $x=0$, provided $y\ne 0$.

The alternative $\phi =y$, $\psi =y/x^2$ is valid on patches with $xy\ne 0$. On $y=0$, it is a parabolic partial differential equation away from the fully degenerate origin, so two independent hyperbolic characteristic coordinates are unavailable.

Parametrize the data by $(\xi ,\eta ,\zeta )=(\cos \theta ,\sin \theta ,-1)$. The determinant testing Noncharacteristic Cauchy data isThus the open upper and lower semicircles are noncharacteristic, but the entire circle is not a valid noncharacteristic data curve: the determinant vanishes at $(1,0)$ and $(-1,0)$.

There is a stronger obstruction at those two points. On $y=0$, the differential equation forces $0=-z^4/3$, and hence $z=0$. This contradicts the prescribed value $z=-1$ at both endpoints. ThereforeThe obstruction is already present if the equation is imposed next to the circle and the solution and its first partial derivatives extend continuously to it.

Away from the endpoints, the characteristic projections are radial half-lines:For $\sin \theta >0$, $r$ increases with $s$; for $\sin \theta <0$, $r$ decreases with $s$. Each projection meets its own open semicircle exactly once and never crosses $y=0$. The endpoint projections themselves are single stationary points, although their lifted characteristic curves have varying height.

A characteristic curve in $(x,y,z)$ follows the vector field defining the first-order partial differential equation. Along such a curve the chain rule turns the equation intoIts characteristic projection is the curve $(x(s),y(s))$ in the independent-variable plane.

Writing $x(0)=\xi$, $y(0)=\eta$, $z(0)=\zeta$, separation of variables givesThese formulas hold on the maximal interval through $s=0$ where the denominators remain nonzero, using the real cube root. They include $\eta =0$ (a stationary characteristic projection) and $\zeta =0$ (the zero height). For $\zeta \ne 0$, the last formula follows from $d(z^{-3})/ds=1$. The alternative zero solution must be retained when deriving this by division by $z^4$.