Past exam of the differential equations and applications course of the University of Oxford / 2008 / B / 5

The square-root reduction of a quadratic eikonal equation explains why two initial normal derivatives are possible. Set $F=p^2+q^2-4u$. Charpit's equations are $\dot{x}=2p$, $\dot{y}=2q$, $\dot{u}=2p^2+2q^2=8u$, $\dot{p}=4p$, $\dot{q}=4q$. Parametrize the initial semicircle by $(x_0,y_0)=(\cos s,\sin s)$, $-\pi /2<s<\pi /2$. Tangential compatibility gives $-p_0\sin s+q_0\cos s=0$, and $F=0$ gives $p_0^2+q_0^2=4$. Thus there are two normal-gradient choices, $p_0=2\epsilon \cos s$, $q_0=2\epsilon \sin s$, $\epsilon =\pm 1$.

Integration gives the requested parametric solutionsFor $\epsilon =+1$, $r=\sqrt{x^2+y^2}=e^{4t}$, so $u_{+}=r^2$ on the whole right half-plane. For $\epsilon =-1$, before focusing at the origin, $r=2-e^{4t}$, so $u_{-}=(2-r{)}^2$ on $x>0$, $0<r<2$. The origin is a caustic for this inward family. The circle $r=2$ is approached only as characteristic time tends to negative infinity, where $u$ and its gradient vanish. It cannot be crossed by this initial characteristic chart. Thus the initial data alone permit two branches; each is locally unique only after its normal-gradient sign is chosen.

On the added half-line write $x_0=a\le 0$, $y_0=1$, $u_0=1$. Tangential differentiation gives $p_0=0$ and $q_0=2\delta$, $\delta =\pm 1$. Its characteristics giveThe plus branch is $u=y^2$ on $x\le 0,y>0$; the minus branch is $u=(2-y{)}^2$ on $x\le 0,y<2$. On the positive $y$-axis, the semicircle traces are $y^2$ and $(2-y{)}^2$, respectively. Therefore only matching sign choices are continuous throughout their common positive-axis coverage. Mixed choices agree merely at the initial junction $y=1$.

For the matching-plus branch the data-covered open domain isFor the matching-minus branch it isIts formula is $(2-\sqrt{x^2+y^2}{)}^2$ in the right portion and $(2-y{)}^2$ in the left portion. The origin is excluded. The negative $y$-axis must also be a cut: the two formulas have different traces there, so including it would destroy continuity. Both matching solutions are $C^1$ across the included positive-axis seam; generally their second transverse derivatives differ, which is compatible with this first-order problem. Values beyond these characteristic charts can sometimes be continued, but are not selected by the initial data through a nonsingular characteristic graph.