Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 1 / 29A (Partial Differential Equations)

For the first-order quasilinear equation $a(x,u)\cdot \nabla u=b(x,u)$, suppose the coefficients and initial data $u=g$ on a smooth hypersurface $\Sigma$ are $C^1$. At an initial point $x_0$, require $a(x_0,g(x_0))\cdot n(x_0)\ne 0$ for a normal $n$ to $\Sigma$; this is precisely the condition that $\Sigma$ be a non-characteristic hypersurface. Then in a neighborhood of $x_0$ there is a unique $C^1$ solution taking the given data. Smooth coefficients and data give corresponding higher regularity.

The method of characteristics solves $\dot{x}=a(x,u)$, $\dot{u}=b(x,u)$ from each initial point. The non-characteristic condition makes the map from the initial coordinates and characteristic time to $x$ locally invertible, so these characteristic solutions define the asserted function.

Characteristics satisfy $dx/ds=1$, $dy/ds=x$, $du/ds=0$. From $(\xi ,0)$ they are $x=\xi +s$, $y=\xi s+s^2/2$, $u=f(\xi )$, so $x^2-2y={\xi }^2$. The initial parameter is recovered on the two initial half-axes byThis is $C^1$ on the displayed open set, takes the prescribed data at $y=0,x\ne 0$, and substitution verifies $u_x+x{u}_y=0$. Each branch may extend separately beyond this chosen domain, but arbitrary data on both half-axes need not allow one common extension.

The initial line has normal $(0,1)$; the transport field is $(1,x)$, whose normal component is $x$. Thus it is non-characteristic exactly away from the origin. At the origin the equation itself would require $u_x(0,0)=f^{\prime }(0)=0$. For generic $f$ this fails, so a $C^1$ solution on a neighborhood of the entire initial line is not guaranteed. The parabola $y=x^2/2$ is the envelope where the recovery of $\xi$ degenerates. Characteristics can also meet both initial half-axes, imposing compatibility such as $f(\xi )=f(-\xi )$ on an attempted larger domain.

Here $dy/ds=1$, $dx/ds=u^2$, $du/ds=0$, so from initial parameter $\xi$ we get $y=s$, $x=\xi +{\xi }^{2}y$ and $u=\xi$. Solving for the branch agreeing with $u=x$ at $y=0$, and rationalizing the apparent division by $y$, givesThis formula is smooth on that open set, which contains every point of the initial line. It satisfies $x=u+y{u}^2$ and $1+2yu=\sqrt{1+4xy}>0$. Differentiation gives $u_x=1/(1+2yu)$, $u_y=-u^2/(1+2yu)$, verifying the equation and the data.

The transport field is $(u^2,1)$, with normal component one on $y=0$, so the initial line is everywhere non-characteristic. This explains why neighborhoods cover the entire initial line, although a single fixed-width strip need not work for unbounded $x$.