Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 105 / 4 / a

For each $t$, letA weak energy solution of a variable-coefficient wave equation is a function $u\in H^1(U_T)$ with zero lateral trace, equivalently $u\in L^2(0,T;H_0^1(U))$, whose time trace satisfies $u(0)={\psi }_0$ in $L^2(U)$ and for whichfor every $v\in H^1(U_T)$ with zero lateral trace and $v(T)=0$. Here $u\in H^1(0,T;L^2(U))$ has a continuous $L^2$ representative, so the displacement trace is well defined.

The velocity condition is encoded by the boundary term in time; an arbitrary space-time $H^1$ function need not have an $L^2$ trace of $u_t$. The equation implies $u_{tt}=f-L(t)u\in L^2(0,T;H^{-1}(U))$, hence $u_t$ has a continuous $H^{-1}$ representative and $u_t(0)={\psi }_1$ in that sense. This is the weak formulation obtained by integration by parts once in time and once in space.

Take the difference of two weak solutions, so $f={\psi }_0={\psi }_1=0$. At the stated $H^1(U_T)$ regularity, it is not justified simply to test by $u_t$, which need not belong to the spatial test space $H_0^1$. Instead use the antiderivative test underlying uniqueness of Sobolev weak solutions of a wave equation.

Fix $s\in (0,T]$ and defineThese are Bochner integrals in $H_0^1$. The function $v$ is an admissible space-time $H^1$ test, $v_t=-u$ before $s$, and $v(t)=z(s)-z(t)$. The time term in the weak formulation isbecause $u(0)=0$. Symmetry of $a^{ij}$ and integration by parts in time giveFor the lower-order part, integration by parts in space givesBounded coefficients, the Poincaré inequality for $v$, and the Young inequality now implySince $Dv(t)=Dz(s)-Dz(t)$,For $s$ in a sufficiently short fixed interval, absorb the last term into the left-hand side. The Gronwall inequality applied to $\Vert u(s){\Vert }_2^2+(\theta /2)\Vert Dz(s){\Vert }_2^2$ proves $u=0$ on that interval. The equation gives continuity of $u_t$ in $H^{-1}$, so both initial traces at its endpoint are again zero. Repeating with the same uniform coefficient bounds covers $[0,T]$. Therefore

For $F(x,t)=t-\tau (x)$, the conormal to its zero set is $dF=(-D\tau ,1)$. The principal symbol of ${\partial }_t^2+L$ isA characteristic hypersurface requires $p(x,t;dF)=0$ on the surface. Thus $\tau$ satisfies the anisotropic eikonal equationThe first-order and zeroth-order terms do not affect this condition.