Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 29C (Partial Differential Equations)

For a first-order quasilinear partial differential equation, take $C^1$ coefficients $a_j,b$, $C^1$ prescribed data $u_0$ on a smooth $C^2$ hypersurface $S$, and a point $x_0\in S$. The data are non-characteristic there ifwhere $n$ is any nonzero normal to $S$. Then there is a neighborhood of $x_0$ on which a unique $C^1$ solution realizes the data. Uniqueness is local, after restricting to a neighborhood swept out by the nearby characteristics.

The method of characteristics solves $\dot{x}=a(x,u)$, $\dot{u}=b(x,u)$ from points of $S$. In local surface coordinates $\eta$, the map $(\eta ,s)\mapsto x(\eta ,s)$ has nonsingular derivative at $s=0$ exactly under the stated normal condition. The inverse function theorem makes these variables local coordinates, giving the asserted solution and uniqueness.

Here the characteristic curves are straight lines $x=y+sa$, $y\cdot n=0$, along which $\dot{u}=c\cdot (y+sa)+d$. If $a\cdot n\ne 0$, every nearby point determines a unique starting point and travel parameter:Integrating along the Characteristic curve givesThe non-characteristic condition is exactly what permits division by $a\cdot n$ and ensures transversality to the data hyperplane. If $a\cdot n=0$, the Characteristic curve lines starting there never leave it. Moreover the prescribed data must satisfy $a\cdot {\nabla }_S{u}_0=c\cdot x+d$ on the hyperplane even to be compatible. If they do, the Characteristic curve construction still leaves off-hyperplane data undetermined; if they do not, no $C^1$ solution exists.

Start a characteristic curve at $(0,\eta )$ with $u=-\eta$. The Characteristic curve equations give $u=-\eta$, $y=\eta +s$, $x=-\eta s$, so $x={\eta }^2-\eta y$. Therefore the two branches areEach is $C^1$ on $D=\{y^2+4x>0\}$. The plus branch matches $u(0,y)=-y$ for $y>0$, whereas the minus branch matches it for $y<0$. Differentiating $u^2+yu-x=0$ gives $u_x=1/(2u+y)$ and $u_y=-u/(2u+y)$, verifying $u_y+u{u}_x=0$ wherever $2u+y\ne 0$.

On the initial line, the normal Characteristic curve speed is $u=-y$, nonzero away from the origin. Thus the local existence theorem gives precisely these separate local branches near positive and negative initial points. At $y^2+4x=0$ the branches merge and their derivatives become singular. There is no $C^1$ solution realizing the entire initial line in a neighborhood of the origin: the data force $u_y(0,0)=-1$, while the equation and $u(0,0)=0$ force $u_y(0,0)=0$. The failure at this Characteristic curve point is a genuine limitation of the printed global initial-data request.