Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 105 / 3

For $s\ge 0$ and $\xi \in \mathbb{R}$, define the characteristic curve $X(t;s,\xi )$ byThe bounded derivative $F_x$ makes $F(t,\cdot )$ globally Lipschitz, uniformly in $t$. On each finite time interval, $|F(t,x)|\le |F(t,0)|+L|x|$, so Gronwall inequality prevents finite-time escape. The Picard-Lindelof theorem therefore gives a unique trajectory for every finite $t\ge 0$. Differentiation in $\xi$ givesso the characteristic flow map is a $C^1$ increasing diffeomorphism.

Along a characteristic, the chain rule changes the equation intoTracing $(t,x)$ backward by the flow therefore givesThe regularity of the flow makes this a $C^1$ classical solution. Conversely, every classical solution obeys the same ordinary differential equation along every characteristic, so the formula also proves uniqueness.

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For $u_0\in L^{\infty }(\mathbb{R})$, call $u\in L^{\infty }({\mathbb{R}}_{+}\times \mathbb{R})$ a weak solution when, for every $\phi \in C_c^1([0,\infty )\times \mathbb{R})$,This follows by multiplying the equation by a test function and applying integration by parts in time and space; $(F\phi {)}_x$ includes the term $F_x\phi$ because the original transport operator is not in divergence form.

If $u$ is $C^1$, test functions supported away from $t=0$ show that $u_t+F{u}_x=g$ as a distributional identity, hence pointwise. Integrating this pointwise equation by parts in the weak identity leavesArbitrary boundary test functions and the fundamental lemma of the calculus of variations give $u(0,x)=u_0(x)$. Thus a $C^1$ weak solution is the unique classical solution from part a.

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With $g=0$, pull the bounded measurable initial value back along the characteristic flow map:The flow is measurable and invertible, so $u$ is measurable and $\Vert u{\Vert }_{\infty }\le \Vert u_0{\Vert }_{\infty }$. Choose smooth $u_0^{(k)}$ converging to $u_0$ in $L_{\mathrm{loc}}^1$ with uniformly bounded essential suprema, and definePart a makes each $u^{(k)}$ a classical, hence weak, solution. On every compact subset of spacetime, the $C^1$ change-of-variables formula for the flow and its locally bounded Jacobian determinant give $u^{(k)}\to u$ in $L^1$. Passing to the limit in the weak identity by dominated convergence proves that $u$ is a bounded weak solution with initial datum $u_0$.

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The characteristic curves are $x=\xi +at$. Along one of them, $U(t)=u(t,\xi +at)$ obeys the separable ordinary differential equationThereforeFor $0\le t<1$, the denominator is positive everywhere. At $t=1$, it first vanishes where $\cos \xi =1$, namely at $\xi =2\pi k$. The solution consequently has finite-time blowup at timealong the points $x=a+2\pi k$. No finite classical solution can continue through that time.

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Write the Inviscid Burgers equation in conservation form asA bounded function $u$ is a weak solution with initial datum $u_0$ whenfor every compactly supported $C^1$ test function $\phi$.

Across a straight discontinuity $x=\sigma t$, integration by parts on its two sides shows that the boundary terms cancel exactly when the Rankine-Hugoniot condition holds:when $u_{-}\ne u_{+}$. For every $c>0$, defineThe three jumps have left and right states $(0,-c)$, $(-c,c)$, and $(c,0)$, so their Rankine-Hugoniot speeds are respectively $-c/2$, $0$, and $c/2$, exactly the speeds of the displayed lines. Hence each $u_c$ satisfies the weak equation away from the origin and across every jump. Moreover, its nonzero support at time $t$ has length $ct$, so $u_c(t,\cdot )\to 0$ in $L_{\mathrm{loc}}^1$ as $t\downarrow 0$; its initial datum is therefore zero in the weak identity.

The zero function and all the distinct functions $u_c$ are bounded weak solutions with the same zero initial datum. Thus weak solutions are not unique. The central jump from $-c$ to $c$ is an expansion shock, which an entropy condition would exclude.

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