Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 105 / 3

The function $f$ is real analytic at $y$ if there is a neighborhood of $y$ on which its multivariable Taylor seriesconverges to $f(x)$. Here $\alpha$ is a multi-index; equivalently, $f$ agrees locally with a convergent real power series centred at $y$.

Locally write the real analytic hypersurface as $\Sigma =\{F=0\}$ with $dF\ne 0$. The conormal bundle is spanned by $dF$. The surface is a characteristic hypersurface at $p$ precisely when the principal symbol vanishes on that conormal:Only the matrix $(a_{ij}+a_{ji})/2$, which is a symmetric matrix, contributes to this expression.

One prescribes analytic Cauchy data: the value of $u$ and one first derivative transverse to $\Sigma$, for examplewhere $\varphi$ and $\psi$ are real analytic on $\Sigma$. Being a non-characteristic hypersurface allows the equation to solve for the second derivative in the transverse direction. The Cauchy-Kovalevskaya theorem then gives one and only one local real analytic solution near $p$.

Expanding the equation givesso its principal symbol isIf a characteristic curve is locally a graph $y=y(x)$, its conormal is proportional to $(-y^{\prime },1)$. The characteristic equation is thereforeOn each region separated by $y=\pm 1$, separation of variables givesThus all the characteristic curves areThe inner curves approach the horizontal characteristics $y=\pm 1$ as $x\to \pm \infty$; the outer hyperbolic-cotangent branches have vertical asymptotes and also approach $y=\pm 1$. This describes the requested sketch.

The initial line $x=0$ has conormal $(1,0)$, and $p(y,1,0)=1$, so it is a non-characteristic hypersurface at every point. Since the coefficients and prescribed data are real analytic, the Cauchy-Kovalevskaya theorem gives a unique real analytic solution in a neighborhood of each point of $\{x=0\}$, hence in a neighborhood of that line.

Put $a(y)=(1-y^2{)}^2$. Multiply $u_{xx}={\partial }_y(a{u}_y)$ by $2u_x$ and integrate over $-1<y<1$. An integration by parts givesbecause $a(1)=a(-1)=0$. Therefore the energy estimate is in fact the conservation lawIf $u(0,y)=u_x(0,y)=0$, then differentiating the first identity in $y$ also gives $u_y(0,y)=0$, so the conserved nonnegative energy is zero. Hence $u_x=0$ and $(1-y^2)u_y=0$ throughout the open strip. There $|y|<1$, so both derivatives vanish; connectedness and the initial value now give

Assume $0<\epsilon <1$ and setThe travel-time coordinate for a one-dimensional variable-speed wave equationsends the initial interval $I_{\epsilon }=(-a,a)$ to $(-S_{\epsilon },S_{\epsilon })$. By the characteristic curves for speed one minus y squared, the two characteristic coordinates are $s-x$ and $s+x$. The finite propagation speed and uniqueness theorem for hyperbolic partial differential equations therefore give the maximal characteristic diamondEquivalently,In the $(x,s)$ plane this is a diamond with vertices $(0,\pm S_{\epsilon })$ and $(\pm S_{\epsilon },0)$; transforming back bends its four sides into the characteristic curves found in part 3(c). Beyond any one of those sides, a point's backward characteristics meet $x=0$ outside $I_{\epsilon }$, where no Cauchy data were prescribed, so uniqueness cannot be extended farther.