Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 5 / 1 / 3

For the scalar conservation law, put $v_0(\xi )=f^{\prime }(u_0(\xi ))$ and $m={\min }_{\xi }{v}_0^{\prime }(\xi )={\min }_{\xi }{f}^{\prime \prime }(u_0(\xi ))u_0^{\prime }(\xi )$. The concave-flux characteristic lifespan and characteristic flow map areFor $t<t^{\ast }$, $X_t^{\prime }=1+t{v}_0^{\prime }>0$, and $X_t(\xi )=\xi +ct$ outside the support of $u_0$. Thus $X_t$ is a global smooth diffeomorphism and the formula is $u(t,x)=u_0(X_t^{-1}(x))$. The method of characteristics proves existence and uniqueness among smooth solutions.

If $m<0$, at a minimizer ${\xi }_{\ast }$ the numerator $u_0^{\prime }({\xi }_{\ast })$ is nonzero, since its product with $f^{\prime \prime }(u_0({\xi }_{\ast }))$ is negative. Thereforeblows up as $t\uparrow t^{\ast }$, showing that this is the maximal smooth lifespan. Strict concavity alone does not require $f^{\prime \prime }$ to be negative at every point; the argument uses no such extra hypothesis.

The inverse function theorem applied to $X_t$ makes $u_0\circ X_t^{-1}$ smooth. Since $u_0(\pm A)=0$, $X_t(\pm A)=\pm A+ct$. Monotonicity of the characteristic flow map and the vanishing of $u_0$ outside $[-A,A]$ implyThus the solution is a smooth function with compact support at each such time. The fact that some interior characteristic speeds differ from $c$ causes no difficulty: they cannot cross the two exterior characteristic curves before $t^{\ast }$.

The compact support just established and $f(0)=0$ permit differentiation of the antiderivative and integration of the scalar conservation law from $-\infty$:This is a Hamilton-Jacobi equation. There is no arbitrary function of time: the normalization at $-\infty$ and the zero exterior flux determine it.

The tangent-line inequality for a differentiable concave function is $f(r)\le f(s)+f^{\prime }(s)(r-s)$. Taking $r=u=U_x$ and using the Hamilton-Jacobi equation yieldsAlong $x(\tau )=x_0+\tau f^{\prime }(s)$, the chain rule identifies the left side with $dU(\tau ,x(\tau ))/d\tau$. Integrating givesStrict concavity makes equality possible exactly when $u=s$ along the line, which will select the maximizing characteristic curve.

Because $f^{\prime }$ is a decreasing bijection, its inverse $h$ exists and is continuous. For any $y$, choose $s=h((x-y)/t)$ in the preceding inequality. The line from $(0,y)$ then reaches $(t,x)$, giving $U(t,x)\ge U(0,y)+tg((x-y)/t)$.

Now let $x_0=X_t^{-1}(x)$ and $s=u_0(x_0)$. On this characteristic curve, $u=s$, so the tangent-line inequality is an equality throughout. This proves attainment andIn fact the maximizing foot is unique. The concave Legendre dual can be written $g(z)={\inf }_s(zs-f(s))$ and satisfies $g^{\prime }(z)=h(z)$: compare the minimizing values at $z$ and $z+\epsilon$ and use continuity of $h$. This avoids assuming differentiability of $h$, which strict concavity by itself does not guarantee. Differentiating the maximizing expression with respect to $y$ gives $u_0(y)-h((x-y)/t)=0$, hence $X_t(y)=x$ and $y=x_0$. The maximum representation for a concave conservation law therefore reproduces the characteristic solution of a scalar conservation law throughout the smooth lifespan.

The printed linear ordering is false for $z<c$. For example, $f(s)=-s^2/2$, $c=0$, $h(z)=-z$, $k_{-}=-3/4$, $k_{+}=-1/4$ satisfy the derivative hypothesis, but at $z=-1$ the printed lower bound would require $3/4\le 1/2$.

The correct inverse-flux quadratic bounds areIndeed $h(c)=0$ and $h(z)/2={\int }_c^z{h}^{\prime }(r)/2dr$; reversing the integration limits reverses the linear inequalities. Equivalently, for $z\ne c$, $k_{-}\le h(z)/(2(z-c))\le k_{+}$. Since $g(c)=0$ and $g^{\prime }=h$, integration once more gives the quadratic inequalities on both sides of $c$. A useful sign-independent consequence is $|h(z)|\le -2k_{-}|z-c|$.

Write $M=\Vert u_0{\Vert }_{L^1}$. Since $|U(0,y)|\le M$, choosing $y=x-ct$ gives $g((x-y)/t)=g(c)=0$ and ${\max }_y{G}_{t,x}(y)\ge -M$. At the maximizing foot $x_0$, the inverse-flux quadratic bounds giveCombining the two estimates and dividing by $-k_{+}>0$ givesThe last step uses $u=h((x-x_0)/t)$ and the sign-independent bound on $h$. This is square-root decay before characteristic crossing; it is proved only for $0<t<t^{\ast }$. No continuation past characteristic crossing or global-time smoothness is assumed. If $M=0$, the initial function and the solution vanish.