Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 5 / 1 / 2

A characteristic curve has $x-at$ constant. Backtracking reaches the initial line if $x\ge at$ and the inflow boundary otherwise. HenceValues on the dividing characteristic curve are immaterial for bounded weak solutions.

For a classical solution on the closed quadrant, choose $u_0,f\in C^1([0,\infty ))$ and impose the corner compatibility for constant-speed transportThese make the values and both first derivatives agree across $x=at$; each branch solves the transport equation. They are also necessary for a $C^1$ solution up to the initial and inflow boundaries.

Spatial $C^{\infty }$ regularity for every $t\ge 0$, up to $x=0$, holds precisely for smooth data with all matching jets:Indeed the spatial derivatives of order $n$ at $x=at$ are $u_0^{(n)}(0)$ and $(-1/a{)}^n{f}^{(n)}(0)$. Necessity of smooth $u_0$ follows at $t=0$; smoothness of $f$ on any finite interval follows by reading the boundary branch in a spatial slice at a larger time.

For the inflow transport boundary condition, the weak formulation isHere $\phi$ is a smooth compactly supported test function on the closed quadrant. Put $z=x-at$ and $\psi (t,z)=\phi (t,z+at)$. The transformed region is $t>\tau (z):=\max (0,-z/a)$. The formula in the preceding solution is $v(t,z)=b(z)$, with $b(z)=u_0(z)$ for $z\ge 0$ and $b(z)=f(-z/a)$ for $z<0$. Consequently the first integral equals $-\int b(z)\psi (\tau (z),z)dz$. Its positive-$z$ portion cancels the initial integral; the substitution $z=-at$ in its negative-$z$ portion cancels the inflow integral, including the factor $a$. This proves weak existence for arbitrary bounded data without corner compatibility.

For uniqueness, subtract two weak solutions. Interior test functions in these characteristic coordinates give ${\partial }_{t}v=0$ distributionally, so $v(t,z)=b(z)$ on each vertical ray. This follows first on rectangles compactly contained in $t>\tau (z)$ using tensor test functions, and then on the whole region by overlapping rectangles. In the full weak formulation the remaining lower-boundary term is $-\int b(z)\psi (\tau (z),z)dz=0$. Arbitrary smooth traces supported separately on $z>0$ and $z<0$ force $b=0$ there. The single point $z=0$ has zero measure. Thus the two-branch formula is the unique bounded weak solution.