Quasilinear Characteristics and Cauchy Kovalevskaya — Examples

The projected characteristics in the independent-variable plane are $X(t,\xi )=(t,\xi +t)$. Thus their spatial component is $\xi +t$, so $\xi =x-t$, and the Jacobian of $(t,\xi )\mapsto (t,\xi +t)$ is $1$.

Along one such curve, $z^{\mathrm{\prime }}=z(1-z)$ and $z(0)=g(\xi )$; put $q=g(\xi )$ and $D(t)=1+q(e^t-1)$. On $D(t)>0$, direct differentiation gives $z=q{e}^t\mathrm{/}D$ and $z^{\mathrm{\prime }}=q{e}^t(1-q)\mathrm{/}{D}^2=z(1-z)$, with $z(0)=q$. This includes $q=0$ and $q=1$ without division by either. Since $D(0)=1$ and $D^{\mathrm{\prime }}(t)=q{e}^t$ has constant sign, $\{t:D(t)>0\}$ is exactly the interval containing zero on which the denominator is nonzero.

Substitute $\xi =x-t$ to obtain the formula, and characteristic reconstruction verifies the PDE on its stated domain.

The formula gives $X=\xi +t\xi =(1+t)\xi$ and $X_{\xi }=1+t>0$.

Inverting gives $\xi =x\mathrm{/}(1+t)$, while $u(t,X)=u_0(\xi )=\xi$; hence $u(t,x)=x\mathrm{/}(1+t)$.

$X_{\xi }=1-t{\mathrm{sech}}^2\xi$, which first vanishes at $t=1$, $\xi =0$, since $0<{\mathrm{sech}}^2\xi \le 1$.

The Riccati formula gives $u_x(t,X(t,0))=-1\mathrm{/}(1-t)$, which tends to $-\mathrm{\infty }$ as $t\uparrow 1$.

The lifted characteristic is $Z(t,\xi )=-\tanh \xi$ and $X(t,\xi )=\xi -t\tanh \xi$, both defined for every $t\ge 0$.

Yet $X_{\xi }(1,0)=0$, so the projected map has a caustic at $t=1$.

Thus the lifted ODE persists while inverse projection to a graph fails, refuting the statement.

$S_x=a$, so $x{S}_x+(S_x{)}^2=ax+a^2=S$; thus $S$ is a complete integral.

$S_a=x+2a=0$ gives $a=-x\mathrm{/}2$, and $S_{aa}=2$ is invertible.

Substitution gives $u=S(x,-x\mathrm{/}2)=-x^2\mathrm{/}4$; the nondegenerate-envelope lemma makes it a classical solution.

For $x\mathrm{\ne }0$, $Du(x)=x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$, so $\mathrm{\mid }Du(x)\mathrm{\mid }=1$.

Along a unit vector $e$, the directional quotients at zero are $u(te)\mathrm{/}t=1$ for $t>0$ and $u(te)\mathrm{/}t=-1$ for $t<0$.

They have no common limit, so $u$ is not differentiable at zero and cannot be classical there.

Here $F_p=(2p_x,0)=0$ on $F=0$, so $[F_p,D\gamma ]$ has rank $1<2$ for $\gamma (y)=(y,0)$.

Every $u(x,y)=f(y)$ with $f(0)=0$ has $u_x=0$, hence solves $F=0$ and attains the data.

Choosing distinct such $f$ gives distinct local solutions, and step 1.1 identifies the failed rank hypothesis.

The equation is solved for the second normal derivative: $u_{tt}=u_{xx}$, whose right side is analytic in the relevant jet variables.

The normal line is $t$, and the data supply exactly ${\mathrm{\partial }}_t^{0}u{\mathrm{\mid }}_{t=0}=g_0$ and ${\mathrm{\partial }}_{t}u{\mathrm{\mid }}_{t=0}=g_1$, the orders $0$ and $1$ required for order $2$.

The coefficient of $u_{tt}$ is $1$, so $t=0$ is noncharacteristic for this solved normal form.

Since $H_x=0$, the momentum equation gives $\dot{p}=0$, hence $p(t,\xi )=-\xi$.

As $\dot{X}=H_p(p)=p$, integration from $X(0,\xi )=\xi$ gives $X=(1-t)\xi$.

$X_{\xi }=1-t$ is nonzero for $t<1$ and zero at $t=1$, which is the first caustic by the stated convention.

Every derivative of $g$ at $0$ is $0$, while $g(x)>0$ for $x\mathrm{\ne }0$; thus $g$ is $C^{\mathrm{\infty }}$ but cannot equal its Taylor series near $0$.

The chain rule gives $u_t=-g^{\mathrm{\prime }}(x-t)$ and $u_x=g^{\mathrm{\prime }}(x-t)$, hence $u_t+u_x=0$ and $u(0,x)=g(x)$.

Translation carries the flat nonanalytic point from $0$ to $x=t$, so $u$ is nonanalytic there despite being smooth.