Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2011 / 9 / Solution

Introduce a characteristic parameter $s$ and label each point of the initial segment by $r\in [0,1]$. The method of characteristics givesIntegrating gives the parametric solutionThe projected characteristic curves satisfy $y=r(2-x)$, so each lies on a straight line through $(2,0)$. Its branch from the data always has $x<2$; the common vertex is approached only as $s\to -\infty$.

Eliminating $s,r$ using $e^s=2-x$ and $r=y/(2-x)$ yieldsIts derivatives give $(x-2)z_x+y{z}_y=y$, and substitution of $x=1$ gives the prescribed value one.

The Jacobian determinant of $(s,r)\mapsto (x,y)$ is $-e^{2s}\ne 0$, and the inverse just displayed is unique. Hence the data determine the solution exactly onThis is the domain determined by a finite segment under a radial characteristic flow. Its interior is $x<2$, $0<y<2-x$; the two boundary rays come from the endpoints of the initial segment. Every point there lies on exactly one characteristic from the data. Outside this sector, a characteristic either misses the initial segment or lies on the branch $x>2$, which cannot reach it. Thus the formula may be extended beyond this sector, but such an extension is not uniquely specified by the given data.