Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 5 / 1 / 1

A weak solution is an equivalence class $u\in L^{\infty }((0,\infty )\times \mathbb{R})$ satisfying, for every test function $\phi \in C_c^{\infty }([0,\infty )\times \mathbb{R})$,This is integration by parts for the nonconservative transport equation. In particular, the weak formulation contains $c_x\phi$ as well as $c{\phi }_x$. The last integral encodes the initial condition; a pointwise boundary value of an arbitrary Lebesgue space representative is not the definition. Compact support of the test function makes the identity meaningful even if $c$ is unbounded.

The characteristic curves solve $\dot{X}=X$, hence $X(t,\xi )=e^t\xi$. By the chain rule, $d[u(t,X(t,\xi ))]/dt=0$. Pulling back along the characteristic flow map givesFor $u_0\in C^1$, this is a classical solution: $u_t=-e^{-t}x{u}_0^{\prime }(e^{-t}x)$ and $u_x=e^{-t}{u}_0^{\prime }(e^{-t}x)$ cancel in the transport equation, and at $t=0$ the initial condition holds. Conversely, constancy along every characteristic curve forces this formula.

Use the same formula for a measurable representative of $u_0$. Dilation preserves null sets and gives $\Vert u{\Vert }_{L^{\infty }}=\Vert u_0{\Vert }_{L^{\infty }}$. To check the weak formulation, change variables $x=e^{t}y$ and set $\psi (t,y)=e^t\phi (t,e^{t}y)$. ThenThe spacetime integral becomes $\int u_0(y){\psi }_t(t,y)dydt=-\int u_0(y)\phi (0,y)dy$, as required.

For uniqueness, take any bounded weak solution and write $v(t,y)=u(t,e^{t}y)$. Choosing $\phi (t,x)=e^{-t}\psi (t,e^{-t}x)$ in the weak formulation givesFor tensor test functions $\psi (t,y)=\eta (t)\theta (y)$, this says that $\int v(t,y)\theta (y)dy$ is distributionally constant and has value $\int u_0\theta$. A countable dense family of spatial test functions identifies $v=u_0$ almost everywhere. Thus the displayed solution is the unique bounded weak solution.