Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 105 / 3 / 2 / e / Solution

For the stated mollifier, $({\partial }_t+{\partial }_s){\chi }_{\epsilon }=({\partial }_x+{\partial }_y){\chi }_{\epsilon }=0$. Thus the interior doubled integral isOn a bounded state range, $H$ is Lipschitz in each argument, and so is $Q$: away from the diagonal each partial derivative has modulus at most $\sup |f^{\prime }|$, and the function is continuous across the diagonal. Local $L^1$ translation continuity therefore makes this converge to the corresponding single-time, single-space integral. The compact time support is chosen below $T$, with a margin for $s\le t+2\epsilon$.

The initial terms need a separate argument. Since the time kernel is supported at $t-s\in [-2\epsilon ,-\epsilon ]$, $B_v$ is identically zero for $t\ge 0$, whereas $B_u$ samples $v(s,\cdot )$ at $\epsilon \le s\le 2\epsilon$.

The averaged initial trace of an entropy solution follows directly from its inequalities. Test the constant-level entropy for $v$ with $(1-s/\delta {)}_{+}\rho (y)$, using smooth approximations of that temporal cutoff. Boundedness of $v$ and its entropy flux givesApproximate $v_0$ locally in $L^1$ by finitely many constants with a smooth nonnegative partition of unity. Applying this estimate on each partition member and using the triangle inequality makes the limsup of ${\delta }^{-1}{\int }_0^{\delta }\Vert v(s)-v_0{\Vert }_{L^1(K)}ds$ arbitrarily small. This proves its convergence to zero for each compact $K$.

The time kernel has size $O({\epsilon }^{-1})$ and the spatial kernel is a normalized approximate identity. The averaged trace, followed by spatial $L^1$ translation continuity of $v_0$, therefore gives $B_u\to \int |u_0-v_0|\phi (0)$. Its time mass is one, not one-half, because its whole support lies on the positive $s$ side. ConsequentlyThis is the Kato inequality for scalar conservation laws with its initial contribution. Assuming an initial trace solely from interior translation continuity would leave a gap; the entropy argument above supplies it.