Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 1 / 30E (Partial Differential Equations)

The characteristic equations for a transport equation are ${\dot{x}}_1=x_1$, ${\dot{x}}_2=2x_2$, $\dot{u}=5u$. Starting from $(\xi ,1)$ at parameter zero gives $(x_1,x_2)=(\xi e^s,e^{2s})$, $u=e^{5s}g(\xi )$. Eliminating $s,\xi$ givesEvery point of this upper half-plane lies on exactly one characteristic from the initial line, so it is the maximal characteristic domain determined by the data. Neither the line $x_2=0$ nor the lower half-plane is reached in finite characteristic time; extra data would be needed there.

For $g\in C^1$ this is a classical solution, directly verified by differentiation. The printed assumption is only continuity, which in general does not give a classical $C^1$ solution. Under that assumption the same formula is the unique continuous characteristic solution and solves the transport equation distributionally; smoothing $g$ locally and passing to limits verifies the latter interpretation. A classical interpretation needs the additional regularity just stated.

Inside the fan $u=x/t$, one has $u_t=-x/t^2$ and $u{u}_x=x/t^2$, while the constant regions also satisfy the equation. Across $x=0$ and $x=t$, the two traces of both $u$ and its flux $u^2/2$ agree. Integrating by parts separately in the three regions therefore cancels all interface terms and givesfor every smooth compactly supported test function. The initial term is correct because $u(\cdot ,t)\to 1_{\{x>0\}}$ in local $L^1$: the difference is supported in $(0,t)$ and its integral is at most $t$. Thus the fan is a weak solution of the Burgers equation with the required initial data.

Write $[u]=u_r-u_l$ and $[f]=u_r^2/2-u_l^2/2$. Splitting the test-function integral along $x=s(t)$ and integrating by parts gives a remaining interface contributionFor example the normal to the interface in $(x,t)$ coordinates is proportional to $(1,-s^{\prime })$, so the jump in the conserved spacetime flux $(u^2/2,u)$ has normal component $[f]-s^{\prime }[u]$. Since the traces of the test function are arbitrary, the interior weak equation holds if and only if $s^{\prime }[u]=[f]$. At a genuine jump $[u]\ne 0$, this isThis is the Rankine-Hugoniot condition. An initial-value weak formulation also requires the prescribed initial trace; the jump condition alone concerns the interior equation. The source's reference to “(1)” is to the displayed Burgers equation, not a different PDE.