Past exam of the differential equations 1 course of the University of Oxford / 2023 / 2 / b / iii / Solution

Use the compatible open semicircles and set $r=\sqrt{x^2+y^2}$. The characteristic curve through $(\cos \theta ,\sin \theta ,-1)$ hasThe condition $s<1$ is essential: at $s=1$ there is finite-time blowup, so the opposite branch cannot be reached from the data. From $r=(1-s\sin \theta {)}^{-1}$ and $y=r\sin \theta$,Eliminating $s$ yieldsThe ratio inside the cube root is positive precisely on the two data-determined regionsOn $D_{+}$ each point traces to the upper semicircle, and on $D_{-}$ to the lower semicircle, with $s<1$. To check the parameter ranges explicitly, for $\eta =\sin \theta >0$ they are $-\infty <s<1$, whereas for $\eta <0$ they are $1/\eta <s<1$. At the latter lower endpoint the planar projection escapes to infinity. The inverse function theorem and uniqueness for the characteristic ordinary differential equations give a unique smooth solution on each region. Equivalently, each point has the unique foot $(x/r,y/r)$ and parameter $(r-1)/y$.

A direct check is particularly short. For $s=(r-1)/y$, let $\mathcal{D}=xy{\partial }_x+y^2{\partial }_y$. Then $\mathcal{D}r=yr$, $\mathcal{D}y=y^2$, andSince $dz/ds=-z^4/3$, the displayed formula solves the partial differential equation, and $r=1$ gives $z=-1$ on both open semicircles. The finite boundary where $r=1+y$, away from the axis, is the parabola $y=(x^2-1)/2$ and corresponds to blowup. Points on $y=0$ cannot be added to a classical continuation of these regions: where they are approached from a data-determined region, the height behaves like a nonzero multiple of $|y{|}^{1/3}$, or has incompatible limits at the endpoints. The formula is therefore the complete classical answer for the compatible arcs, while the full-circle problem as printed has no classical solution.