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

A weak solution is an equivalence class $u\in L^{\infty }((0,\infty )\times \mathbb{R})$ satisfying, for every test function $\phi \in C_c^{\infty }([0,\infty )\times \mathbb{R})$,This is integration by parts for the nonconservative transport equation. In particular, the weak formulation contains $c_x\phi$ as well as $c{\phi }_x$. The last integral encodes the initial condition; a pointwise boundary value of an arbitrary Lebesgue space representative is not the definition. Compact support of the test function makes the identity meaningful even if $c$ is unbounded.

The characteristic curves solve $\dot{X}=X$, hence $X(t,\xi )=e^t\xi$. By the chain rule, $d[u(t,X(t,\xi ))]/dt=0$. Pulling back along the characteristic flow map givesFor $u_0\in C^1$, this is a classical solution: $u_t=-e^{-t}x{u}_0^{\prime }(e^{-t}x)$ and $u_x=e^{-t}{u}_0^{\prime }(e^{-t}x)$ cancel in the transport equation, and at $t=0$ the initial condition holds. Conversely, constancy along every characteristic curve forces this formula.

Use the same formula for a measurable representative of $u_0$. Dilation preserves null sets and gives $\Vert u{\Vert }_{L^{\infty }}=\Vert u_0{\Vert }_{L^{\infty }}$. To check the weak formulation, change variables $x=e^{t}y$ and set $\psi (t,y)=e^t\phi (t,e^{t}y)$. ThenThe spacetime integral becomes $\int u_0(y){\psi }_t(t,y)dydt=-\int u_0(y)\phi (0,y)dy$, as required.

For uniqueness, take any bounded weak solution and write $v(t,y)=u(t,e^{t}y)$. Choosing $\phi (t,x)=e^{-t}\psi (t,e^{-t}x)$ in the weak formulation givesFor tensor test functions $\psi (t,y)=\eta (t)\theta (y)$, this says that $\int v(t,y)\theta (y)dy$ is distributionally constant and has value $\int u_0\theta$. A countable dense family of spatial test functions identifies $v=u_0$ almost everywhere. Thus the displayed solution is the unique bounded weak solution.

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.

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.