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

The growth bound is $|F(t,x)|\le K(1+|x|)$. A characteristic curve through $(s,a)$ solvesThe $C^1$ field is locally Lipschitz in space, giving local existence and uniqueness. On either direction of a bounded time interval the Gronwall inequality gives $1+|X(t;s,a)|\le (1+|a|)e^{K|t-s|}$. Thus no trajectory can escape to infinity at finite time. On the resulting compact space-time region, local ordinary differential equation existence continues it, proving global definition for all $t\in \mathbb{R}$.

Uniqueness gives the flow composition law and inverse $X(s;t,\cdot )$. Differentiation with respect to $a$ solves the variational equation, making each map a $C^1$ diffeomorphism. Its positive Jacobian determinant isThese facts constitute the global characteristic flow under linear growth. A uniform global bound on $D_{x}F$ is not required; its bounds on each compact characteristic tube suffice.

Write $X(t,a)=X(t;0,a)$ and $A(t,x)=X(0;t,x)$. If $u_0\in C^1({\mathbb{R}}^d)$, the unique global classical solution isIt is $C^1$, and the chain rule makes its value constant along every characteristic curve. The same calculation proves classical uniqueness. No boundedness of $u_0$ is needed for this classical statement.

For $u_0\in L^{\infty }$, a bounded weak solution on the positive half-line is defined byfor every compactly supported $C^1$ test function on $[0,\infty )\times {\mathbb{R}}^d$. The analogous backward-time identity defines the solution on all real time. The $u\mathrm{div}F$ term is essential: this is transport, not the conservative equation $u_t+\mathrm{div}(Fu)=0$.

The same inverse-flow formula defines the unique weak solution, with $\Vert u(t){\Vert }_{\infty }=\Vert u_0{\Vert }_{\infty }$, equality almost everywhere, and a representative continuous into $L^1$ on each compact spatial set. For a direct uniqueness argument, change variables $x=X(t,a)$ in the weak identity and put $w(t,a)=u(t,X(t,a))$. Using $J_t=(\mathrm{div}F)(t,X)J$ givesinitially for $C^1$ compactly supported $\zeta$. Since $J>0$ and $J,J_t$ are continuous on compact sets, approximation in the spatial parameter permits $\zeta =\eta /J$: both $\zeta$ and ${\zeta }_t$ are uniformly approximable there, which are the only derivatives left in this identity. Hence $\int w{\eta }_t+\int u_0\eta (0)=0$ and $w=u_0$ almost everywhere. This also verifies existence by reversing the argument, and avoids assuming second derivatives of the $C^1$ field. Local $L^1$ continuity follows by approximating $u_0$ by continuous functions on compact flow tubes.

This is the weak transport solution under characteristic coordinates. The weak-strong uniqueness principle says that a weak solution in the specified class agrees with a classical solution having the same data for as long as that classical solution exists. Here it holds globally in the bounded class whenever the classical initial data are also bounded.

When the velocity depends on $u$, different initial values travel at different speeds. The resulting characteristic crossing can destroy a classical solution in finite time, even though the initial function is smooth and bounded. For the Inviscid Burgers equation, take $u_0(a)=-\tanh a$. Before crossing,andAt $a=0$ this tends to $-\infty$ as $t\uparrow 1$, while $|u|\le 1$. Thus $\boxed{\text{bounded states can undergo finite-time gradient blow-up and shock formation}.}$ The bounded-derivative Burgers flux agrees with this equation on the whole state range, if a global bounded-speed flux is desired. After crossing, arbitrary weak solutions need not be unique; an entropy solution condition is needed to select the admissible continuation.

A bounded weak solution satisfiesfor every compactly supported $C^1$ test function, of either sign, allowed to meet $t=0$. Across a moving jump $x=s(t)$ with distinct left/right constant states, the distributional delta coefficient vanishes exactly when the Rankine-Hugoniot condition holds:It is required on every interface of a piecewise constant solution; the initial trace must also agree with the data.

For nonuniqueness respecting the paper's global bound on $f^{\prime }$, use the bounded-derivative Burgers flux $f(z)=z^2/2$ on $|z|\le 1$ and $f(z)=|z|-1/2$ outside. It is $C^1$ and $|f^{\prime }|\le 1$. Take data $-1$ for $x<0$, $1$ for $x>0$. A stationary jump obeys the condition because $f(-1)=f(1)$. Another weak solution is the rarefaction waveThe fan solves the equation inside each smooth region, is continuous at its edges, and has the same local $L^1$ initial trace. Thus weak solutions are not unique. The stationary expansion shock violates the entropy inequality, whereas the fan is admissible; using the unmodified quadratic flux alone would miss the printed global bounded-derivative hypothesis.

For a bounded $C^1$ classical solution and a smooth convex entropy $\eta$, let its entropy flux for a scalar conservation law satisfy ${\psi }^{\prime }=f^{\prime }{\eta }^{\prime }$. The chain rule givesIntegration against a nonnegative test function, including its initial boundary contribution, gives equality in the entropy inequality. Approximate a convex piecewise $C^1$ entropy by smooth convex functions on the bounded state range. Their values and flux antiderivatives converge uniformly, so the integrated equality passes to the limit. Therefore every bounded classical solution is an entropy solution. The bounded qualifier is required by the definition's $L^{\infty }$ membership; an arbitrary unbounded $C^1$ solution would not satisfy that membership condition.

Conversely, the affine entropies ${\eta }_{+}(z)=z$ and ${\eta }_{-}(z)=-z$ are both convex, with fluxes $f$ and $-f$. Their inequalities force the same linear weak integral to be both nonnegative and nonpositive. It is therefore zero for every nonnegative test, hence for every test by subtraction of nonnegative compactly supported tests. Thus $\boxed{\text{entropy solution}⟹\text{weak solution}.}$ No assumption that the flux itself is convex is used.

Since the derivative of $|z-k|$ is $\mathrm{sgn}(z-k)$ away from $k$, choose the Kruzhkov entropy fluxOn either side of $k$ its derivative is $f^{\prime }(z){\eta }_k^{\prime }(z)$. The factor $f(z)-f(k)$ vanishes at the change of sign, so the function is continuous and there is no extra delta contribution to its weak derivative. Adding a constant to the flux changes none of the test integrals. Substitution into the entropy inequality pairs $|u-k|$ with this flux, not with an arbitrary antiderivative unrelated to the chosen entropy.

Set $H(a,b)=|a-b|$ and $Q(a,b)=\mathrm{sgn}(a-b)(f(a)-f(b))$. Both are symmetric in $a,b$. For fixed $(s,y)$ apply the entropy inequality for $u$ at the constant level $k=v(s,y)$ with test $\Phi (t,x,s,y)$, then integrate over $(s,y)$. This producesHere the first integral is over both time-space pairs and the second over the three remaining variables. Repeat with $v$, level $u(t,x)$, integrating over $(t,x)$, to obtain the analogous terms with ${\Phi }_s,{\Phi }_y$ and initial value $v_0$.

Adding and using symmetry giveswhere $B_u=\int H(u_0(x),v(s,y))\Phi (0,x,s,y)dsdxdy$ and $B_v=\int H(u(t,x),v_0(y))\Phi (t,x,0,y)dtdxdy$. All integrals are finite on the test's compact support because the states are bounded. The levels being inserted are constants relative to the variables in each entropy inequality; there is no differentiation of the merely measurable other solution. Fubini's theorem and continuity of $H,Q$ justify the integration. This is doubling of variables for scalar conservation laws.

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.

Put $h=|u-v|$, $q=Q(u,v)$ and $h_0=|u_0-v_0|$. The mean value theorem gives $|q|\le Mh$ on the bounded state range. Fix a final time $t$ and let $L(\tau )=a-M(t-\tau )$ and $R(\tau )=b+M(t-\tau )$. Choose a smooth increasing function $\rho$ equal to zero on $(-\infty ,0]$ and one on $[1,\infty )$. For small $\delta >0$, defineIts two nonnegative derivative contributions imply $(W_{\delta }{)}_{\tau }+M|(W_{\delta }{)}_x|\le 0$. Therefore $h(W_{\delta }{)}_{\tau }+q(W_{\delta }{)}_x\le 0$.

Use the nonnegative test $\phi (\tau ,x)={\eta }_{ϵ}(\tau )W_{\delta }(\tau ,x)$, where ${\eta }_{ϵ}(0)=1$, it is one up to just before $s$, decreases smoothly to zero near $s$, and remains zero afterward. The Kato inequality for scalar conservation laws givesbecause the other interior contribution is nonpositive. At Lebesgue times $s$, the temporal approximate identity tends to $\int h(s)W_{\delta }(s)$; take a countable sequence $\delta \downarrow 0$ and use dominated convergence on the bounded cone. This yieldsfor almost every $s\in [0,t]$. This is local L1 contraction for scalar conservation laws. It uses only local integrability, so the bounded data need not have finite global $L^1$ norm. If $M=0$, the flux difference is zero and the same proof uses a stationary interval.

If $u_0=v_0$, the right-hand side of the local L1 contraction for scalar conservation laws estimate is zero. Applying it on a countable collection of rational intervals and final times proves $u=v$ almost everywhere in space-time. Thus

To deduce nonnegativity one needs a one-sided comparison, rather than merely putting $v=0$ in the absolute-value estimate. The difference $w=u-v$ satisfies the weak equation with flux $f(u)-f(v)$. Subtract its weak integral identity from the Kato inequality for scalar conservation laws and divide by two. The result is the same inequality forWhen $w\ge 0$ both expressions vanish; when $w<0$, $|q_{-}|\le M{w}_{-}$. The shrinking-interval proof therefore bounds the negative part by its initial negative part. This is order preservation for scalar entropy solutions.

The constant $v=0$ is an entropy solution for every $f$, even if $f(0)\ne 0$, because the constant flux has zero spatial derivative. If $u_0\ge 0$, its initial negative part is zero, and comparison givesThis explicitly establishes the additional sign conclusion without incorrectly identifying uniqueness alone with order preservation.