Scalar Conservation Laws and Entropy Solutions — Examples

The speed is the chord slope. By [F3], $f$ is $C^2$ and strictly convex, and the given states satisfy $u_L>u_R$. The Riemann solver [F1] therefore supplies a weak entropy shock with speed $s=(f(u_R)-f(u_L))\mathrm{/}(u_R-u_L)$. Since $u_R-u_L\mathrm{\ne }0$, algebra gives $s=\frac{1}{2}(u_R^2-u_L^2)\mathrm{/}(u_R-u_L)=\frac{1}{2}(u_L+u_R)$, so this is exactly the profile and speed in the statement.

Admissibility. With $f^{\mathrm{\prime }}(u)=u$, the Lax inequalities read $u_R=f^{\mathrm{\prime }}(u_R)\le s\le f^{\mathrm{\prime }}(u_L)=u_L$, and indeed $u_R\le \frac{1}{2}(u_L+u_R)\le u_L$ because $u_R\le u_L$; the jump is compressive and entropy-admissible by [F2]. Alternatively the chord through $(u_R,\frac{1}{2}{u}_R^2)$ and $(u_L,\frac{1}{2}{u}_L^2)$ lies above the parabola, which is the chord criterion of [F2].

Conclusion via the solver and the initial trace. By [F1] the shock with speed $s$ is the unique Kruzhkov entropy solution of the Riemann problem, so the weak conservation law, all Kruzhkov entropy inequalities and the strong local $L^1$ trace hold. The trace can also be seen directly: the set where $u(t,\cdot )$ differs from the step datum $u_0$ is contained in the interval between $0$ and $st$, of length $\mathrm{\mid }s\mathrm{\mid }t$ and amplitude $\mathrm{\mid }{u}_L-u_R\mathrm{\mid }$, so its $L^1$ discrepancy on any compact set is at most $\mathrm{\mid }{u}_L-u_R\mathrm{\mid }\mathrm{\mid }s\mathrm{\mid }t\to 0$. For $u_L=1$, $u_R=0$ the formula gives $s=\frac{1}{2}$, so the shock is the ray $x=t\mathrm{/}2$, with state $1$ on the left and $0$ on the right.

Specialisation of the solver. By [F2], $f^{\mathrm{\prime }}(u)=u$ and $(f^{\mathrm{\prime }}{)}^{-1}(\xi )=\xi$; the three branches of the strictly convex Riemann solver [F1] read $u_L$ for $x\mathrm{/}t\le u_L$, $x\mathrm{/}t$ for $u_L<x\mathrm{/}t<u_R$, and $u_R$ for $x\mathrm{/}t\ge u_R$, which is exactly the displayed centred rarefaction. Hence by [F1] it is the unique Kruzhkov entropy solution of the Riemann problem, satisfies the weak conservation law, all Kruzhkov entropy inequalities, and attains the Riemann datum in the strong local $L^1$ sense.

Direct check of the middle branch and the interfaces. On the fan, $u(t,x)=x\mathrm{/}t$, so $u_t=-x\mathrm{/}{t}^2$, $u_x=1\mathrm{/}t$ and $u_t+u{u}_x=-x\mathrm{/}{t}^2+(x\mathrm{/}t)(1\mathrm{/}t)=0$ by the chain rule [F3]; on the two outer regions $u$ is constant, so both derivatives vanish there. At $x=u_{L}t$ the three-branch formula gives $u_L$ from both the first and middle branches, and at $x=u_{R}t$ it gives $u_R$ from both the middle and last branches; the traces of $u$ and of $f(u)$ therefore agree across the rays, so by [F3] no interface terms arise in the weak residual and the profile is a distributional weak solution.

Entropy production and regularity. For any convex $C^2$ pair $(\eta ,q)$ with $q^{\mathrm{\prime }}={\eta }^{\mathrm{\prime }}{f}^{\mathrm{\prime }}$, the chain rule gives ${\mathrm{\partial }}_t\eta (u)+{\mathrm{\partial }}_{x}q(u)={\eta }^{\mathrm{\prime }}(u)(u_t+f^{\mathrm{\prime }}(u)u_x)={\eta }^{\mathrm{\prime }}(u)(u_t+u{u}_x)$ on each smooth branch; this vanishes on the fan by step 1.2 and on the outer branches because $u$ is constant. Across the rays the traces of $\eta (u)$ and $q(u)$ agree because $u$ is continuous there, so no interface measure arises: the entropy production is identically $0$ on $\mathbb{R}\times (0,\mathrm{\infty })$ for every convex $C^2$ pair. (For the non-smooth Kruzhkov pairs, the entropy inequalities are supplied by the solver [F1].) On the fan $u_x=1\mathrm{/}t$, so the profile is locally Lipschitz on every compact subset of the open strip $t>0$; no uniform Lipschitz bound as $t\downarrow 0$ is claimed.

Initial trace and the special case. The discrepancy from the initial step is supported between $\min \{0,u_L\}t$ and $\max \{0,u_R\}t$, and is bounded by $u_R-u_L$. Thus ${\int }_K\mathrm{\mid }u(t)-u_0\mathrm{\mid }\le (u_R-u_L)(\max \{0,u_R\}-\min \{0,u_L\})t\to 0$. When $u_L=0$, $u_R=1$, the fan is $x\mathrm{/}t$ on $0<x<t$, with outer states $0$ and $1$. For nonsmooth convex pairs, smooth convex approximation and uniform convergence of the integral fluxes pass the zero-production identity of step 2.1 to the limit; thus production is zero, not merely nonpositive.

The data. $u_0(y)=-\arctan y$ is smooth and bounded, with $u_0^{\mathrm{\prime }}(y)=-1\mathrm{/}(1+y^2)\in [-1,0)$ and ${\min }_y{u}_0^{\mathrm{\prime }}=-1$ attained only at $y=0$. The flux $f(u)=\frac{1}{2}{u}^2$ is $C^{\mathrm{\infty }}$ with $f^{\mathrm{\prime }}(u)=u$.

The characteristic map is a diffeomorphism for $t<1$. $X_t^{\mathrm{\prime }}(y)=1+t{u}_0^{\mathrm{\prime }}(y)=1-\frac{t}{1+y^2}=\frac{1+y^2-t}{1+y^2}>0$ for $0\le t<1$ and all $y$, so by the mean value theorem [F2] $X_t$ is strictly increasing; moreover $X_t(y)=y+O(1)$ tends to $\pm \mathrm{\infty }$ as $y\to \pm \mathrm{\infty }$, so $X_t$ maps $\mathbb{R}$ onto $\mathbb{R}$. A strictly increasing surjection is a homeomorphism, and since $X_t^{\mathrm{\prime }}$ never vanishes, the inverse function theorem [F2] makes the inverse $y(\cdot ,t)=X_t^{-1}$ smooth with $y_x(x,t)=1\mathrm{/}{X}_t^{\mathrm{\prime }}(y(x,t))$ and, differentiating $X_t(y(x,t))=x$ in $t$, $y_t(x,t)=-u_0(y(x,t))\mathrm{/}{X}_t^{\mathrm{\prime }}(y(x,t))$.

The outer branches for $t>1$. The function $h(a)=t\arctan a-a$ increases up to $\sqrt{t-1}$ and then decreases to $-\mathrm{\infty }$, so its unique positive zero $a(t)$ satisfies $a(t)>\sqrt{t-1}$. Hence $X_t^{\mathrm{\prime }}>0$ on $[a(t),\mathrm{\infty })$, and $X_t$ maps this interval bijectively onto $[0,\mathrm{\infty })$; by oddness it maps $(-\mathrm{\infty },-a(t)]$ bijectively onto $(-\mathrm{\infty },0]$. For $x>0$ choose the unique $y>a(t)$ with $X_t(y)=x$, and for $x<0$ choose the unique $y<-a(t)$; define $u(x,t)=-\arctan y$. These branches are smooth by the inverse function theorem, solve Burgers directly: implicit differentiation gives $y_x=1\mathrm{/}{X}_t^{\mathrm{\prime }}(y)$, $y_t=-u_0(y)\mathrm{/}{X}_t^{\mathrm{\prime }}(y)$, hence $u_t+u{u}_x=0$, and have traces $u^{-}=\arctan a(t)$, $u^{+}=-\arctan a(t)$ at $x=0$. Their fluxes agree, so the stationary jump satisfies Rankine--Hugoniot and is entropy-admissible by [F3].

The ansatz is a classical solution. Put $u(x,t)=u_0(y(x,t))$ for $0\le t<1$, which is smooth in $(x,t)$. By the chain rule and step 1.2, $u_t=u_0^{\mathrm{\prime }}(y)y_t=-u_0^{\mathrm{\prime }}(y)u_0(y)\mathrm{/}{X}_t^{\mathrm{\prime }}(y)$ and $u_x=u_0^{\mathrm{\prime }}(y)y_x=u_0^{\mathrm{\prime }}(y)\mathrm{/}{X}_t^{\mathrm{\prime }}(y)$. Hence $u_t+f(u{)}_x=u_t+f^{\mathrm{\prime }}(u)u_x=u_0^{\mathrm{\prime }}(y)[-u_0(y)+u_0(y)]\mathrm{/}{X}_t^{\mathrm{\prime }}(y)=0$, so $u$ solves $u_t+f(u{)}_x=0$ classically on $\mathbb{R}\times (0,1)$. Since $X_t$ satisfies ${\dot{X}}_t(y)=u_0(y)=f^{\mathrm{\prime }}(u(X_t(y),t))$, this is exactly the family of characteristics of [F1], along which $u$ is the constant $u_0(y)$.

The gradient formula. Differentiating $u(X_t(y),t)=u_0(y)$ in $y$ and using $u_x{X}_t^{\mathrm{\prime }}=u_0^{\mathrm{\prime }}$ gives $u_x(X_t(y),t)={\displaystyle \frac{u_0^{\mathrm{\prime }}(y)}{X_t^{\mathrm{\prime }}(y)}}={\displaystyle \frac{u_0^{\mathrm{\prime }}(y)}{1+t{u}_0^{\mathrm{\prime }}(y)}}=-{\displaystyle \frac{1}{1+y^2-t}}$. At $y=0$, where $X_t(0)=0$, this reads $u_x(0,t)=-1\mathrm{/}(1-t)$ for $t<1$.

Catastrophe at $T_{\ast }=1$. For each $t<1$, ${\sup }_x\mathrm{\mid }{u}_x(x,t)\mathrm{\mid }={\sup }_y\frac{1}{1+y^2-t}=\frac{1}{1-t}\to \mathrm{\infty }$ as $t\uparrow 1$, the supremum being attained at $y=0$; the classical solution exists for every $t<1$ by step 2.1, and its slope becomes unbounded as $t\uparrow 1$. Hence the first gradient catastrophe occurs at $T_{\ast }=1$.

The limit profile at $t=1$. $X_1^{\mathrm{\prime }}(y)=\frac{y^2}{1+y^2}\ge 0$ with equality only at $y=0$, so $X_1(y)=y-\arctan y$ is strictly increasing with range $\mathbb{R}$; its inverse is continuous, and the profile $u(x,1)=u_0(X_1^{-1}(x))$ is continuous. For $y\mathrm{\ne }0$, implicit differentiation as in step 2.2 with $t=1$ gives $u_x(X_1(y),1)=-1\mathrm{/}{y}^2$, which tends to $-\mathrm{\infty }$ as $y\to 0$, i.e. as the corresponding point $x=X_1(y)\to 0$. Thus at $t=1$ the solution is still continuous but has unbounded slope at $x=0$: a gradient catastrophe, not a jump.

Entropy continuation and uniqueness. The branches of step 1.3 give a bounded piecewise smooth profile for $t>1$. Its only jump is the descending stationary shock, so graph integration and the chord criterion give the weak equation and all smooth convex entropy inequalities; smooth convex approximation gives the Kruzhkov inequalities. For $t<1$ the smooth solution of step 2.1 has zero entropy production and attains $u_0$ locally uniformly. As $t\to 1$ from either side, the selected feet converge for every $x\mathrm{\ne }0$ to $X_1^{-1}(x)$; boundedness and dominated convergence give matching local $L^1$ traces to the continuous profile of step 3.2. Integrating separately below and above $t=1$ and taking these traces cancels the time-interface terms in both weak and entropy pairings. Thus this is a global entropy solution with the stated datum. Its uniqueness follows from [F3], even though $-\arctan x$ is not integrable. The first slope blow-up is at $t=1$, and a nonzero shock is present for every $t>1$.

The initial trace. For a compact $K\subseteq {\mathbb{R}}^n$ and $0<t<T$, the two definitions of $u(t,x)$ and $u_0(x)$ differ exactly on $\{x\in K:x\cdot \eta \text{ lies strictly between }0\text{ and }ct\}$, a set of measure at most $C_K\mathrm{\mid }t\mathrm{\mid }$; hence ${\int }_K\mathrm{\mid }u(t,x)-u_0(x)\mathrm{\mid }dx\le \mathrm{\mid }{u}_R-u_L\mathrm{\mid }{C}_K\mathrm{\mid }t\mathrm{\mid }\to 0$ as $t\downarrow 0$. So $u$ has the strong local $L^1$ trace $u_0$.

The interface computation. Choose $k$ with ${\eta }_k\mathrm{\ne }0$, and put $\gamma (t,x^{\mathrm{\prime }})=(ct-{\sum }_{j\mathrm{\ne }k}{\eta }_j{x}_j)\mathrm{/}{\eta }_k$. For a test supported in $t>0$, integrate first in $x_k$ on each side of $x_k=\gamma$. FTC and the moving-endpoint rule give the weak pairing $\frac{1}{\mathrm{\mid }{\eta }_k\mathrm{\mid }}{\int }_{{\mathbb{R}}^{n-1}\times (0,T)}\phi (t,x^{\mathrm{\prime }},\gamma (t,x^{\mathrm{\prime }}))(c[u]-[f]\cdot \eta )d{x}^{\mathrm{\prime }}dt\mathrm{.}$ For ${\eta }_k>0$ the lower side is the left state; for ${\eta }_k<0$ the lower side is the right state, which reverses the jump and converts ${\eta }_k$ to $\mathrm{\mid }{\eta }_k\mathrm{\mid }$. The endpoint derivatives are ${\gamma }_t=c\mathrm{/}{\eta }_k$ and ${\gamma }_{x_j}=-{\eta }_j\mathrm{/}{\eta }_k$.

The initial boundary. For a test meeting $t=0$, perform the same integration on $\delta <t<T$. The bottom term is $-\int u(\delta ,x)\phi (\delta ,x)dx$. By step 1.1 it converges to $-\int u_0\phi (0,x)dx$, cancelling the prescribed initial term. Thus the full Cauchy residual is the interface integral of step 1.2 over $0<t<T$.

Necessity. If $c(u_R-u_L)-(f(u_R)-f(u_L))\cdot \eta \mathrm{\ne }0$, then it is nonzero on a small interface patch; by [F3] there is a nonnegative smooth compactly supported test function supported in a small space--time neighbourhood of a point of that patch and positive on the patch, and step 2.1 makes the weak residual for this test nonzero, contradicting the weak identity.

Sufficiency and normal form. Conversely, if $(f(u_R)-f(u_L))\cdot \eta =c(u_R-u_L)$, the interface bracket vanishes identically and step 2.1 shows that the weak identity holds for every test function; the trace was verified in step 1.1, so $u$ is a distributional weak solution. Since $\eta$ is a unit vector, ${\mathrm{\nabla }}_{t,x}(x\cdot \eta -ct)=(-c,\eta )$ has length $\sqrt{1+c^2}$, so the unit normal from the minus side is $\nu =(-c,\eta )\mathrm{/}\sqrt{1+c^2}$ and the condition becomes $[u]{\nu }_t+[f(u)]\cdot {\nu }_x=(-c[u]+[f(u)]\cdot \eta )\mathrm{/}\sqrt{1+c^2}=0$. No division by the jump was used at any point.

The production measure. On $\{x<st\}$ the profile equals the constant $u_L$ and on $\{x>st\}$ it equals $u_R$; for a piecewise constant function with a single jump of speed $s$ the distributional derivatives are the jump measures described in [F2]. Hence ${\mathrm{\partial }}_t{\eta }_k(u)+{\mathrm{\partial }}_x{q}_k(u)=([q_k]-s[{\eta }_k])\delta (x-st)$ with $[q_k]=q_k(u_R)-q_k(u_L)$ and $[{\eta }_k]={\eta }_k(u_R)-{\eta }_k(u_L)$; positive coefficients violate the entropy inequality.

The cases $k\le u_R$ and $k\ge u_L$. If $k\le u_R$, then both states lie above $k$, so ${\eta }_k(u_L)=u_L-k$, ${\eta }_k(u_R)=u_R-k$, $q_k(u_L)=\frac{u_L^2-k^2}{2}$, $q_k(u_R)=\frac{u_R^2-k^2}{2}$ by [F3], and $[q_k]-s[{\eta }_k]=\frac{u_R^2-u_L^2}{2}-\frac{u_L+u_R}{2}(u_R-u_L)=0$. If $k\ge u_L$, then both states lie below $k$, so ${\eta }_k(u)=k-u$ and $q_k(u)=\frac{k^2-u^2}{2}$ at both states by [F3]; hence $[q_k]=\frac{u_L^2-u_R^2}{2}$ and $[{\eta }_k]=u_L-u_R$, and the same subtraction again gives $0$.