Past exam of the differential equations 1 course of the University of Oxford / 2021 / 1 / a

The method of characteristics replaces the first-order partial differential equation by the ordinary differential equationsAn integrating factor for the second equation gives $d(e^{-2s}y)/ds=\beta C{e}^{-s}$, so their general solution of a differential equation isHere $C,D,E$ are real constants. For $C\ne 0$, eliminating $s$ gives the characteristic projectionseach restricted to either $x>0$ or $x<0$. These are arcs of parabolas, with the straight-line exception $K=0$. Along an oriented characteristic curve, $|x|$ increases with $s$.

For $C=0$, the characteristic projections lie on $x=0$: $y=D{e}^{2s}$ describes either vertical half-line when $D\ne 0$. For $C=D=0$ the projection is the single point $(0,0)$, although $u=2s+E$ still changes. This last lifted characteristic curve cannot be contained in the function graph of a classical solution: at $(0,0)$ the original partial differential equation would require $0=2$.