= Solution
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, $p_0=2\varepsilon\cos s$, $q_0=2\varepsilon\sin s$, $\varepsilon=\pm1$.
<Integration> gives the requested parametric solutions
$$
\boxed{\begin{aligned}x&=[1+\varepsilon(e^{4t}-1)]\cos s,\\y&=[1+\varepsilon(e^{4t}-1)]\sin s,\\u&=e^{8t},\quad p=2\varepsilon e^{4t}\cos s,\quad q=2\varepsilon e^{4t}\sin s.\end{aligned}}
$$
For $\varepsilon=+1$, $r=\sqrt{x^2+y^2}=e^{4t}$, so $u_+=r^2$ on the whole right half-plane. For $\varepsilon=-1$, 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 $x_0=a\leq0$, $y_0=1$, $u_0=1$. Tangential <differentiation> gives $p_0=0$ and $q_0=2\delta$, $\delta=\pm1$. Its <characteristics> give
$$
\boxed{x=a,\qquad y=1+\delta(e^{4t}-1),\qquad u=e^{8t}.}
$$
The plus branch is $u=y^2$ on $x\leq0,y>0$; the minus branch is $u=(2-y)^2$ on $x\leq0,y<2$. On the positive $y$-axis, the semicircle traces are $y^2$ and $(2-y)^2$, respectively. Therefore \b[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 is
$$
\boxed{D_+=\{x>0\}\cup\{y>0\},\qquad
u_+=\begin{cases}x^2+y^2,&x>0,\\y^2,&x\leq0,\ y>0.\end{cases}}
$$
For the matching-minus branch it is
$$
\boxed{D_-=\{x>0,\ x^2+y^2<4\}\cup\{x<0,\ y<2\}\cup\{x=0,\ 0<y<2\}.}
$$
Its 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.
\Image[/past-exam-of-the-differential-equations-and-applications-course-of-the-university-of-oxford/2008-charpit-domains.png]
{title=Data-covered domains for the two matching-sign characteristic solutions}
Back to article page
