Past exam of the differential equations and applications course of the University of Oxford / 2011 / B / 6

Write $F_p,F_q$ evaluated at the fixed pair $(p(s),q(s))$. Differentiating in $\tau$ gives $x_{\tau }=F_p$, $y_{\tau }=F_q$, and $u_{\tau }=p{F}_p+q{F}_q$, proving $u_{\tau }=p{x}_{\tau }+q{y}_{\tau }$.

Differentiate the other parameter. The difference of the required sides isThe first bracket is zero by the prescribed initial compatibility condition, and the second is zero by differentiating $F(p(s),q(s))=0$. Thus $u_s=p{x}_s+q{y}_s$ as well. These calculations use the differentiability of the supplied parametric functions.

If the coordinate Jacobian determinant $J=x_s{y}_{\tau }-y_s{x}_{\tau }$ is nonzero, the inverse function theorem gives local coordinates $(x,y)$. The chain rule then makes $(u_x-p,u_y-q)$ solve a homogeneous two-by-two linear system with determinant $J$. ThereforeThis is the method of characteristics for a first-order equation depending only on the gradient.

Take $F(p,q)=p^2+pq-1$. Parametrize the initial line by $x_0=s>0$, $y_0=0$, $u_0=2s^2$. Initial compatibility gives $p=4s$, and $F=0$ then gives $q=1/(4s)-4s$. The characteristic formula in part (a) yieldsSince $y>0$, take $\tau >0$. To select the branch connected smoothly to the initial data, stop each characteristic before its coordinate map folds. Its Jacobian determinant isso this branch has $0<\tau <2s^2$. Eliminating $\tau =y/(4s)$ givesFor fixed $y>0$, the right side defining $x-y$ is strictly increasing on that selected branch. At its lower endpoint it equals $3y^{1/3}/4$, and it tends to infinity as $s\to \infty$. Thus it gives exactly one $s$ for every point inSubstituting that unique $s$ into the formula for $u$ gives the required classical solution. As $y\downarrow 0$ at fixed $x>0$, $s\to x$, and $u\to 2x^2$ as required.

Without the pre-fold branch restriction, the algebraic equation also has a smaller positive root. That root tends to zero as $y\downarrow 0$, and its $u$ tends to zero rather than $2x^2$, so it does not solve the prescribed initial-boundary problem. The fold $J=0$ is the characteristic caustic; uniqueness here is for the classical branch continued from the given initial data before that fold.