Past exam of the differential equations and applications course of the University of Oxford / 2009 / B / 4 / Solution

For the general equation the Charpit's equations areAt parameter time zero set $(x,y,u)=(x_0,y_0,u_0)$ and choose $p_0,q_0$ to satisfy $F(x_0,y_0,u_0,p_0,q_0)=0$ and $p_0{x}_0^{\prime }+q_0{y}_0^{\prime }=u_0^{\prime }$. The chain rule givesso $F$ remains zero. The non-characteristic condition is $x_0^{\prime }{F}_q-y_0^{\prime }{F}_p\ne 0$, ensuring a local graph.

For the particular equation put $F=p^3+(y-1)q+u-2$ and label the initial curve by $y_0=1+s$, $0<s\le 1$, $x_0=1$, $u_0=2+2s^3$. Compatibility gives $q_0=6s^2$ and $p_0^3=-8s^3$, hence the unique real choice $p_0=-2s$. Integrating the characteristic equations givesThe projection determinant is $J=12s^2{e}^t(1-2e^{-2t})$. Initially $J=-12s^2\ne 0$; its first zero is $t=\frac{1}{2}\log 2$. At that fold,To describe the entire chart attached to the data before the fold, put $Y=y-1>0$ and $w=e^{-2t}>1/2$. Elimination gives $x-1=6Y^{2}w(1-w)$ andThe initial-endpoint constraint $s\le 1$ therefore gives the uniquely determined chartThe boundary $s=1$ is included as inherited endpoint data; open-domain statements use strict inequality there. The fold is excluded. For forward characteristic times only, $0\le t<\frac{1}{2}\log 2$, impose additionally $x\ge 1$. The source excludes $s=0$, so $(1,1)$ is not a non-characteristic initial point. The sheet beyond the fold does not extend this initial graph uniquely.