Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 1 / 29A (Partial Differential Equations)

The characteristics solve ${\dot{x}}_j=a_j(x)$ and $\dot{u}=b(x,u)$, starting with $x$ on the initial hypersurface and $u=\phi (x)$. The non-characteristic hypersurface condition at a point is $a\cdot n\ne 0$, where $n$ is any nonzero normal there. It makes the map from initial-surface coordinates and characteristic time to $x$ locally invertible. Local uniqueness for the ordinary differential equations and the inverse function theorem then give a unique local $C^1$ solution near that point, for the stated $C^1$ coefficients and data.

Along the characteristics, $x_1+x_2$ is constant. Since $d{x}_2/dt=-1$, the amplitude equation gives $du/d{x}_2=-x_{2}u$. Matching the initial value at $x_2=0$ yieldsDifferentiation directly verifies both the partial differential equation and the initial data.

The solution is constant on characteristics $x_1+x_2=\eta$. On the initial curve this invariant is $h(s)=s$ for $s<0$ and $h(s)=s+s^2$ for $s\ge 0$. This is a $C^1$ bijection of $\mathbb{R}$, with everywhere positive derivative. Its inverse isTherefore $\boxed{u(x_1,x_2)=f(h^{-1}(x_1+x_2)).}$ The inverse is $C^1$ even at zero, so for $C^1$ data this is a global $C^1$ solution. Every characteristic meets the initial curve exactly once.

The new characteristics have invariant $x_1-x_2$. On the initial curve it is $D(s)=s$ for $s<0$ and $D(s)=s-s^2$ for $s\ge 0$. The latter branch folds at $s=1/2$, attaining its maximum $1/4$. Some characteristics meet the initial curve twice or more; generic values of $f$ at those intersections disagree, so no global solution constant along characteristics can exist.

At every initial point except $s=1/2$, the curve is non-characteristic, and the local existence and uniqueness theorem applies. At $s=1/2$, tangency requires $f^{\prime }(1/2)=0$, but this is not sufficient. The fold tangency obstruction for characteristic data requires locally $f(s)=F(D(s))$ with $F$ continuously differentiable: equal invariant values must have equal data, and $f^{\prime }(s)/(1-2s)$ must have a finite continuous limit. For example $f(s)=|s-1/2{|}^{3/2}$ satisfies $f^{\prime }(1/2)=0$ and agrees across the fold, but would force an infinite derivative of $F$ at $1/4$.

If the full compatibility condition holds, $F$ can be extended to invariant values above $1/4$ to give a local solution. Those values are not specified by the initial curve, so local uniqueness then fails.