Past exam of the differential equations and applications course of the University of Oxford / 2010 / B / 4

Use the initial-curve parameter $s$ and a characteristic parameter $t$. Charpit's equations areAt $t=0$ prescribe $x=x_0(s)$, $y=y_0(s)$ and $u=u_0(s)$. The two initial derivatives $p_0,q_0$ must satisfyThe second equation is the chain rule along the initial curve. For the usual local Cauchy construction also require the non-characteristic condition $x_0^{\prime }{F}_q-y_0^{\prime }{F}_p\ne 0$. It ensures both local determination of an allowed branch of $p_0,q_0$ and local inversion of the projected characteristic coordinates.

Along a characteristic, the chain rule and the five equations giveSince $F=0$ initially, $\boxed{F\equiv 0}$ along each constructed curve. With the contact compatibility $du=pdx+qdy$ propagated by this system, an invertible characteristic chart reconstructs $p=u_x,q=u_y$ and hence a local classical solution.

Set $F=xp+q^2/4+u-1$. Parametrize the data by $x_0=s$, $y_0=0$, $u_0=1-s^2/3$. Tangential differentiation gives $p_0=-2s/3$; substitution into $F=0$ then gives $q_0^2=4s^2$. Thus the data allow two normal-derivative branches, indexed by $\epsilon =\pm 1$, with $q_0=2\epsilon s$. For $s>0$ each branch is non-characteristic; the endpoint $s=0$ is degenerate.

The Charpit's equations reduce to $\dot{x}=x$, $\dot{y}=q/2$, $\dot{p}=-2p$, $\dot{q}=-q$ and $\dot{u}=xp+q^2/2$. Integrating with these initial values givesDirect substitution checks $xp+q^2/4+u-1=0$, and $t=0$ recovers the initial data. The projection onto the $(x,y)$ plane has Jacobian determinantIt first vanishes at $t=\log 2$. These points form the characteristic caustic $y=\epsilon x/4$, $0<x\le 2$; a single classical solution cannot in general be continued through the fold by treating all projected characteristic curves as a graph.

For a precise description put $r=\epsilon y$. On the chart containing the initial curve and lying before the fold, $t<\log 2$, elimination givesThe plus square-root is forced by $s/x>1/2$, rather than a free second choice. The conditions $0<s\le 1$ give the complete unfolded chartInterior points of the domain have strict inequality at the boundary generated by $s=1$ as well. The other boundary is the characteristic caustic; the original endpoint $s=0$ contributes no open chart. On the portion obtained by following characteristics forward from the data, $0\le t<\log 2$, the region is more specificallyThe $s=1$ boundary is $y=\epsilon (1-1/x)$ and meets the characteristic caustic at $(2,\epsilon /2)$. Negative characteristic times give the continuation across the initial segment into the opposite half-plane. Each sign branch is a distinct local solution; the initial values alone do not select one of them.

The solid shading follows $t\ge 0$; the lighter shading shows the same smooth chart at $t<0$. Reflected panels correspond to the two allowed signs of the initial normal derivative. Beyond the dashed characteristic caustic the characteristic parametrization no longer gives this smooth graph.