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

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.

Expanding the equation givesFor the hyperbolic partial differential equation discriminant, $A=1$, $B=\rho$, $C={\rho }^2-1$, soFor a curve $\rho =\rho (t)$, the characteristic hypersurface equation is $({\rho }^{\prime }{)}^2-2\rho {\rho }^{\prime }+{\rho }^2-1=0$, or ${\rho }^{\prime }=\rho \pm 1$. Therefore the two families of characteristic curves areThe values $C=0$ include the characteristic lines $\rho =-1$ and $\rho =1$. Every curve in the first family approaches $\rho =-1$ as $t\to -\infty$; every curve in the second approaches $\rho =1$. The two curves through $(\rho ,t)=(0,0)$ are $\rho =e^t-1$ and $\rho =1-e^t$, shown as the thick curves below. Equivalently, the characteristic coordinates are $e^{-t}(\rho +1)$ and $e^{-t}(\rho -1)$.

The all-time conclusion in the original PDF is false. The expanding-flow transformation of a wave equation makes both the correct forward statement and a counterexample transparent. PutFor $D={\partial }_t+\rho {\partial }_{\rho }$, the expanded operator from part (i) is $D^2+D-{\partial }_{\rho }^2$. In the new variables $D=-S{\partial }_S$ and ${\partial }_{\rho }=S{\partial }_X$, henceThus it is the ordinary wave equation on the half-plane $S>0$. By the D'Alembert formula, any $C^2$ solution has the formAt $t=0$, differentiating the zero displacement with respect to $\rho$ and using the zero velocity gives, for $-1<\rho <1$,Subtracting $(\rho -1)$ times the first equation from the second gives $2F^{\prime }(\rho +1)=0$, and then $G^{\prime }(\rho -1)=0$. Thus $F=c$ on $(0,2)$ and $G=-c$ on $(-2,0)$. If $t\ge 0$ and $-1<\rho <1$, both arguments remain in those intervals. Consequently the valid forward result isFor negative time the arguments need not stay in the initial intervals. Take $F(s)=(s-2{)}_{+}^3$ and $G=0$, where $q_{+}=\max \{q,0\}$. Thenis a global $C^2$ solution, has the prescribed zero initial displacement and velocity on $(-1,1)$, but $u(0,-\log 3)=1$. This directly disproves the assertion for $t\in \mathbb{R}$.

The connection with part (i) is directional finite propagation speed. At a future point in the strip, the two characteristics traced back to $t=0$ meet it at ${\rho }_0=e^{-t}(\rho +1)-1$ and ${\rho }_0=e^{-t}(\rho -1)+1$, both in $(-1,1)$. For negative time these intersections can lie outside that interval. The lines $\rho =\pm 1$ are characteristic barriers for this forward domain of dependence, not barriers in both time directions.