Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 7 / 1

The characteristic equations for a transport equation are $\dot{X}=V$ and $\dot{V}=X$, so $\ddot{X}=X$. For an initial point $(x_0,v_0)$ their solution isThis is the hyperbolic characteristic flow for an inverted oscillator. The addition formulas give $A_s{A}_t=A_{s+t}$ and $A_t^{-1}=A_{-t}$. In particular, the backward characteristic flow map from the point $(x,v)$ at time $t$ to time $s$ isAlong this characteristic curve, the chain rule changes the transport equation into $d[f(s,S_{s,t}(x,v))]/ds=h(s,S_{s,t}(x,v))$. Integrating from zero to $t$ givesThe assumed $C^1$ regularity makes this a classical solution: on every compact set, the integrand and its needed derivatives are continuous, so differentiation under the finite-time integral is justified. At $t=0$ it has the required initial value, and the characteristic calculation verifies the equation. Conversely every classical solution must satisfy the same integrated identity, proving uniqueness. This is the Duhamel formula for Hamiltonian transport, with Hamiltonian $(v^2-x^2)/2$.

The derivative of the backward characteristic flow map isThus the change of variables formula preserves phase-space Lebesgue measure. With zero source, $f_t=f_0\circ S_{0,t}$, so for every finite $p>0$,Taking the $p$th root proves $\Vert f_t{\Vert }_p=\Vert f_0{\Vert }_p$. For $0<p<1$ this is a quasi-norm, and the argument still works because it uses only a change of variables, not the triangle inequality. The identity also holds in the extended sense when an integral is infinite. Since the flow is bijective and measure-preserving, it additionally preserves the essential supremum, so the same conclusion holds for $p=\infty$.

Taking the essential supremum in the characteristic solution gives the sharper estimateIndeed, the bijective characteristic flow map preserves each spatial-velocity essential supremum. If $H_t={\mathrm{esssup}}_{0\le s\le t,x,v\in \mathbb{R}}|h(s,x,v)|$, this proves $\Vert f_t{\Vert }_{\infty }\le \Vert f_0{\Vert }_{\infty }+t{H}_t$. A time-independent source has $H_t=\Vert h{\Vert }_{L^{\infty }({\mathbb{R}}^2)}$, which is the displayed form. For a time-dependent source, the same symbol must mean a bound uniform over the elapsed time interval; its norm at the final time alone need not bound the accumulated forcing.

The bound is sharp. Take $f_0=1$ and $h=1$, giving $f(t,x,v)=1+t$ and equality for every $t\ge 0$. Both functions are smooth and bounded, although their finite-$p$ integrals over the whole plane are infinite.

Choose $f_0(x,v)=e^{-(x^2+v^2)}$ and $h(t,x,v)=x^2+v^2$. The initial Gaussian function is smooth and belongs to every finite Lp space, and is also bounded. The source is smooth and nonzero. The accumulated source along the backward characteristic curve isIts quadratic-form eigenvalues are $(e^{2t}-1)/2$ and $(1-e^{-2t})/2$. Both are positive for $t>0$, soConsequently $\Vert f_t{\Vert }_p=\infty$ for every $t>0$ and every finite $p\ge 1$; it is unbounded, so its $L^{\infty }$ norm is infinite as well. This spatially nonintegrable forcing in transport example avoids any ambiguity about whether the last endpoint is included in “all $p$”.

For each $p\ge 1$ with $f_0\in L^p$, a sufficient condition is $h\in L^1([0,T];L^p({\mathbb{R}}^2))$ on every finite interval. The Minkowski integral inequality and measure preservation give the finite-time Lp bound for Hamiltonian transportFor all finite $p$ simultaneously, impose, for example, $h\in L_{\mathrm{loc}}^1([0,\infty );L^1\cap L^{\infty })$. The elementary bound $\Vert h_s{\Vert }_p\le \Vert h_s{\Vert }_1^{1/p}\Vert h_s{\Vert }_{\infty }^{1-1/p}$ and the Holder inequality in time show that its $L^p$ norms are locally integrable for every $1\le p<\infty$. If a bounded initial value is also required, the same assumption controls the $p=\infty$ endpoint by the previous part.

A concrete stronger condition, compatible with a nonzero $C^1$ source, is that on each finite time interval the source has a common compact support in $(x,v)$. Its continuity then makes it bounded on that compact cylinder, so all these integrability conditions hold. A nonzero smooth source compactly supported in phase space supplies examples.