Existence of classical solutions. Consider the conservation law \[\begin{aligned} u_t + u^3 u_x = 0 \quad & \textrm{for } (x,t) \in \mathbb{R} \times (0,\infty), \\ u(x,0) = u_0(x) \quad & \textrm{for } x \in \mathbb{R}, \end{aligned}\] where \[u_0(x) = \left\{ \begin{array}{cl} 1 & x<0, \\ (1-x^2)^2 & 0 \le x \le 1, \\ 0 & x > 1. \end{array} \right.\] We have \(c(u)=u^3\) and \[c(u_0(s)) = \left\{ \begin{array}{cl} 1 & s<0, \\ (1-s^2)^6 & 0 \le s \le 1, \\ 0 & s > 1. \end{array} \right.\] Note that \(c(u_0(s))\) is decreasing for \(s \in (0,1)\). By Theorem 3.5, the conservation law has a classical solution \(u:\mathbb{R} \times [0,t_{\mathrm{c}}) \to \mathbb{R}\) where \[t_{\mathrm{c}} = \inf_{s \in (0,1)} \frac{-1}{c(u_0(s))_s} = \inf_{s \in (0,1)} \frac{1}{12 s (1-s^2)^5}.\] Note that \(1/(12 s (1-s^2)^5) \to \infty\) as \(s \to 0,1\). Therefore its minimum in \((0,1)\) is attained at a critical point. We compute \[\begin{aligned} \frac{d}{ds} \frac{1}{12 s (1-s^2)^5} & =\frac{d}{ds} (12 s(1-s^2)^5)^{-1} \\ & = -(12 s(1-s^2)^5)^{-2} (12(1-s^2)^5 -120s^2(1-s^2)^4) \\ & = \frac{-12 (1-s^2)^4(1-11s^2)}{(12 s(1-s^2)^5)^{2}}. \end{aligned}\] If follows that, for \(s \in (0,1)\), \[\frac{d}{ds}\frac{1}{12 s (1-s^2)^5} = 0 \quad \Longleftrightarrow \quad 1-11s^2 = 0 \quad \Longleftrightarrow \quad s^2 = \frac{1}{11}.\] Therefore the only critical point of \(1/(12 s (1-s^2)^5)\) in \((0,1)\) is \(s= 1/\sqrt{11}\), and this is its global minimum point. Its minimum value is \[t_{\mathrm{c}} = \left. \frac{1}{12 s (1-s^2)^5} \right|_{s=1/\sqrt{11}} = \frac{\sqrt{11}}{12(10/11)^5} = \boxed{\frac{11^5 \sqrt{11}}{10^5 12} \approx 0.445}\]
Breakdown of classical solutions. Consider the conservation law \[\begin{aligned} u_t + u u_x = 0 \quad & \textrm{for } (x,t) \in \mathbb{R} \times (0,\infty), \\ u(x,0) = \frac{1}{1+x^2} \quad & \textrm{for } x \in \mathbb{R}. \end{aligned}\]
We have \(c(u)=u\) and \(c(u_0(s))=u_0(s)=(1+s^2)^{-1}\). Observe that \(c(u_0(s))_s = -2s/(1+s^2)^2 < 0\) for \(s >0\) Therefore \(c(u_0(s))\) is decreasing for \(s>0\) and by Theorem 3.9, \(u_x(x_c(t),t)\) blows up at time \[t_{\mathrm{c}} = \inf_{s \in >0 } \, \frac{-1}{c(u_0(s))_s} = \inf_{s >0} \, \frac{(1+s^2)^2}{2s},\] where \[x_c(t)=s_c+ t c(u_0(s_c)).\] Let us find now \(s_c\). Note that \((1+s^2)^2/(2s) \to \infty\) as \(s \to 0,\infty\). Therefore its minimum in \((0,\infty)\) is attained at a critical point. We compute \[\frac{d}{ds} \frac{(1+s^2)^2}{2s} = \frac{8s^2(1+s^2)-2(1+s^2)^2}{4 s^2} = \frac{(1+s^2)(6s^2-2)}{4 s^2}.\] Therefore, for \(s>0\), \[\frac{d}{ds} \frac{(1+s^2)^2}{2s} = 0 \quad \Longleftrightarrow \quad 6s^2-2 = 0 \quad \Longleftrightarrow \quad s^2 = \frac{1}{3}.\] It follows that the only critical point of \((1+s^2)^2/(2s)\) in \((0,\infty)\) is \(s_c=1/\sqrt{3} \approx 0.5774\), and this is its global minimum point we were looking for. Its minimum value is \[t_{\mathrm{c}} = \left. \frac{(1+s^2)^2}{2s} \right|_{s=1/\sqrt{3}} = \frac{8 \sqrt{3}}{9} = \boxed{\frac{8}{3 \sqrt{3}} \approx 1.5396}\] as required. By the same theorem, we also know that \(u_t(x_c(t),t)\) blows up at \(t_c\), provided \(c(u_0(s_c))\neq 0\). By computing \[c(u_0(s_c)) = \frac{3}{4}\neq 0,\] so, indeed \(u_t(x_c(t),t)\) blows up at \(t_c\).
To conclude, since \(x_c:=x_c(t_c) = \frac{\sqrt{3}}{3}+\frac{8\sqrt{3}}{9}\frac{3}{4}=\sqrt{3}\), we conclude that both derivatives \(u_t\) and \(u_x\) blow up at the point \((x_c,t_c)=(\sqrt{3},\frac{8\sqrt{3}}{9}).\)
The characteristic through \((x,t)=(s,0)\) has equation \[x = s + c(u_0(s))t = s + \frac{t}{1+s^2} \quad \Longleftrightarrow \quad t = (1+s^2)(x-s).\] The slope \(1+s^2\) is decreasing for \(s<0\) and increasing for \(s>0\). Therefore the characteristics diverge for \(s<0\) (Figure 1, blue). For \(s=0\) the characteristic is the line \(t=x\) (Figure 1, red). For \(s>0\) the characteristics converge on each other (Figure 1, green), and they first cross at time \(t_c\).
For \(t \in [0,t_c)\), the Method of Characteristics gives \[u(x,t) = u_0(s) = u_0(x-c(u_0(s))t) = u_0(x-c(u(x,t))t) = \frac{1}{1 + (x-u(x,t)t)^2}.\] Therefore \[\boxed {u(x,t) = \frac{1}{1 + (x-u(x,t)t)^2}}\]
Geometric interpretation of \(t_{\mathrm{c}}\): Crossing characteristics.
(i) Let \(t>t_{\mathrm{c}}\) and \(x=s_{\mathrm{c}}+c(u_0(s_{\mathrm{c}}))t\). Define \(h(s,t)=s+c(u_0(s))t\). Note that \(h(s_{\mathrm{c}},t)=x\) by definition of \(x\). Recall that \[t_{\mathrm{c}} = \frac{-1}{c'(u_0(s_{\mathrm{c}}))u_0'(s_{\mathrm{c}})}.\] We have \[h_s(s,t) = 1+ c(u_0(s))_s t = 1+c'(u_0(s))u_0'(s)t.\] Therefore \[h_s(s_{\mathrm{c}},t) = 1+c'(u_0(s_{\mathrm{c}}))u_0'(s_{\mathrm{c}})t = 1 - \frac{t}{t_{\mathrm{c}}} < 1 - 1 = 0\] because \(t>t_{\mathrm{c}}\). Since \(h(s_{\mathrm{c}},t)=x\) and \(h_s(s_{\mathrm{c}}, t)<0\), then there exists \(s_1>s_{\mathrm{c}}\) such that \(h(s_1,t)<x\). But \(h(s,t) \to \infty\) as \(s \to \infty\). In particular, there exists \(s_2 > s_1\) with \(h(s_2,t)>x\). Since \(h\) is continuous, the Intermediate Value Theorem implies that there exists \(s_* \in (s_1,s_2)\) such that \(h(s_*,t)=x\). In conclusion, \[h(s_{\mathrm{c}},t) = x = h(s_*,t)\] and hence \((x,t)\) lies on the characteristic through \((s_{\mathrm{c}},0)\) and on the characteristic through \((s_*,0)\).
(ii) Now, if we have two crossing characteristics \(x_1(t)=s_1+tc(u_0(s_1))\) and \(x_2(t)=s_2+tc(u_0(s_2))\), crossing at some point \((x,t)\), we have that \(x_1(t)=x_2(t) =x\). This means, after rearranging that \[s_1-s_2 = t\left(c(u_0(s_2))-c(u_0(s_1))\right).\] By the assumptions of the problem, we have that the mapping \(\mathbb{R}\ni s\mapsto c(u_0(s))\) is Lipschitz continuous. Actually, this is also Lipschitz continuous when one restricts the domain to \(I:=\{s\in\mathbb{R}: c'(u_0(s))u'_0(s)<0\}\), so define \[L_I:=\sup_{s\in I}|c'(u_0(s))u'_0(s)|.\] By the previous arguments, one must have \[|s_1-s_2| = t |c(u_0(s_2))-c(u_0(s_1))|\le t L_I |s_1-s_2|,\] which clearly implies that \(t\ge\frac{1}{L_I}\). Now, comparing this to the definition of \(t_c\), i.e. \[t_c:=\inf_{s\in I} - \frac{1}{c'(u_0(s))u_0'(s)} = \frac{1}{L_I},\] thus \(t\ge t_c\).
It is very important to notice that in the previous argument we assumed that \(s_1\) and \(s_2\) are located in the same connected component of \(I\).
Otherwise, let us suppose that \(I\) is disconnected and \(s_1\) and \(s_2\) are located in two different connected components of \(I\). Without loss of generality, we can suppose that \(s_1<s_2\). In this case, we can write again \[|s_1-s_2| = t |c(u_0(s_2))-c(u_0(s_1))| = t |c'(u_0(s_0))u'_0(s_0)| |s_1-s_2|,\] where in the second equation we have used the mean value property for derivatives and \(s_0\in (s_1,s_2)\). Now, if \(s_0\in I\), we are again done by the previous arguments.
Otherwise, we have that \(s_0\notin I\) and so \(c'(u_0(s_0))u'_0(s_0)\ge 0\). In this case if \[0\le c'(u_0(s_0))u'_0(s_0)\le L_I,\] we are again done. So it remains to show that for these crossing characteristics with base points \(s_1,s_2\in I\), one cannot have that \[c'(u_0(s_0))u'_0(s_0)> L_I.\] Indeed, if this would be the case, it would mean that \(c\circ u_0\) can increase way faster than as it is decreasing. Why such a condition is possible, in such a scenario we would necessary need to have that if crossing is occurring after a base point \(s_0\) where the function increases a lot, in the mean value theorem one must have that \(c'(u_0(s_0))u'_0(s_0)<0\) or at least \(c'(u_0(s_0))u'_0(s_0)\le L_I\).
The Fundamental Lemma of the Calculus of Variations. Let \(g: \Omega \to \mathbb{R}\) be a continuous function on an open set \(\Omega \subseteq \mathbb{R}^n\). Suppose that \[\begin{equation} \label{FLotCoV} \int_\Omega g(\boldsymbol{x}) \psi(\boldsymbol{x}) \, d \boldsymbol{x}= 0 \quad \textrm{for all } \, \psi \in C_c^\infty(\Omega). \end{equation}\] We show that \(g=0\). Assume for contradiction that \(g \ne 0\). Without loss of generality, assume that there exists a point \(\boldsymbol{p}\in \Omega\) such that \(g(\boldsymbol{p})>0\). Since \(g\) is continuous, then there exists an open set \(U \subset \Omega\) containing \(\boldsymbol{p}\) such that \(g(\boldsymbol{x})>0\) for all \(\boldsymbol{x}\in U\). Choose \(\varepsilon > 0\) such that \(B_\varepsilon(\boldsymbol{p}) \subset U\), where \(B_\varepsilon(\boldsymbol{p})\) is the open ball of radius \(\varepsilon\) centred at \(\boldsymbol{p}\). Choose any nonzero \(\varphi \in C_c^{\infty}(B_\varepsilon(\boldsymbol{p}))\) such \(\varphi \ge 0\), i.e., \(\varphi(\boldsymbol{x}) \ge 0\) for all \(\boldsymbol{x}\in B_\varepsilon(\boldsymbol{p})\). For example, we could take \[\varphi(\boldsymbol{x}) = \left\{ \begin{array}{cl} \exp \left( - \frac{1}{1-(2 | \boldsymbol{x}- \boldsymbol{p}| / \varepsilon) ^2 } \right) & \textrm{if } |\boldsymbol{x}-\boldsymbol{p}| < \varepsilon/2, \\ 0 & \textrm{if } |\boldsymbol{x}-\boldsymbol{p}| \ge \varepsilon/2. \end{array} \right.\] Then \[\int_\Omega g(\boldsymbol{x}) \varphi(\boldsymbol{x}) \, d \boldsymbol{x} = \int_{B_\varepsilon(\boldsymbol{p})} g(\boldsymbol{x}) \varphi(\boldsymbol{x}) \, d \boldsymbol{x}> 0,\] which contradicts equation \(\eqref{FLotCoV}\). Therefore \(g=0\), as required.
Weak formulation of scalar conservation laws. In order to write down the weak formulation of the conservation law we need to compute a flux function \(f\), which satisfies \(f'(u)=c(u)=\ln|u|+1\). Note that \(f\) is only unique up to a constant. We can choose \[\begin{aligned} f(u) & = \int_0^u (\ln |s| + 1) \, ds = \int_0^u \left( \frac{ds}{ds} \ln |s| + 1 \right) \, ds \\ & = s \ln |s| \Big|_{0}^u - \int_0^u s \, \frac{d}{ds} \ln |s| \, ds + \int_0^u 1 \, ds = u \ln |u| - \int_0^u 1 \, ds + \int_0^u 1 \, ds \\ & = u \ln |u|. \end{aligned}\] Therefore \(u \in L^{\infty}(\mathbb{R} \times (0,\infty))\) is a weak solution of the conservation law if \[\boxed{ \int_{t=0}^\infty \int_{x=-\infty}^\infty [u(x,t) \varphi_t(x,t)+u(x,t) \ln |u(x,t)| \varphi_x (x,t)] \, dx dt + \int_{-\pi}^\pi (1+ \cos x) \varphi(x,0) \, dx = 0 }\] for all \(\varphi \in C_c^\infty(\mathbb{R} \times [0,\infty))\).
Weak formulation of scalar conservation laws in several variables. We will use the following integration by parts formula, which can be proved in the same way as Lemma 3.15: Let \(u:\mathbb{R}^n \times [0,\infty) \to \mathbb{R}\), \(\boldsymbol{v}:\mathbb{R}^n \times [0,\infty) \to \mathbb{R}^n\) be continuously differentiable and let \(\varphi \in C_c^\infty(\mathbb{R}^n \times [0,\infty))\). Then \[\begin{gathered} \int_{0}^\infty \int_{\mathbb{R}^n} [u_t(\boldsymbol{x},t)+\mathrm{div} \, \boldsymbol{v}(\boldsymbol{x},t)] \varphi(\boldsymbol{x},t) \, d\boldsymbol{x}dt = \\ - \int_{\mathbb{R}^n} u(\boldsymbol{x},0) \varphi(\boldsymbol{x},0) \, d \boldsymbol{x} - \int_{0}^\infty \int_{\mathbb{R}^n} [u(\boldsymbol{x},t) \varphi_t(\boldsymbol{x},t) + \boldsymbol{v}(\boldsymbol{x},t) \cdot \nabla \varphi(\boldsymbol{x},t)] \, d\boldsymbol{x}dt. \end{gathered}\]
Let \(u \in C^1(\mathbb{R}^n \times [0,\infty))\) be bounded and satisfy \[\begin{align} \label{Q5:2} u_t + \mathrm{div} \boldsymbol{f}(u) = 0 \quad & \textrm{for } (\boldsymbol{x},t) \in \mathbb{R}^n \times (0,\infty), \\ \label{Q5:3} u(\boldsymbol{x},0) = u_0(\boldsymbol{x}) \quad & \textrm{for } \boldsymbol{x}\in \mathbb{R}^n. \end{align}\] Multiply equation \(\eqref{Q5:2}\) by \(\varphi \in C_c^\infty(\mathbb{R}^n \times [0,\infty))\), integrate over \(\mathbb{R}^n \times [0,\infty)\), and apply the integration by parts formula with \(\boldsymbol{v}(\boldsymbol{x},t)=\boldsymbol{f}(u(x,t))\) to obtain \[\begin{aligned} 0 & = \int_0^\infty \int_{\mathbb{R}^n} [u_t(\boldsymbol{x},t) + \mathrm{div}_{\boldsymbol{x}} \boldsymbol{f}(u(\boldsymbol{x},t))] \varphi(\boldsymbol{x},t) \, d \boldsymbol{x}dt \\ & = - \int_{\mathbb{R}^n} u(\boldsymbol{x},0) \varphi(\boldsymbol{x},0) \, d \boldsymbol{x} - \int_{0}^\infty \int_{\mathbb{R}^n} [u(\boldsymbol{x},t) \varphi_t(\boldsymbol{x},t) + \boldsymbol{f}(u(\boldsymbol{x},t)) \cdot \nabla \varphi(\boldsymbol{x},t)] \, d\boldsymbol{x}dt \\ & = - \int_{\mathbb{R}^n} u_0(\boldsymbol{x}) \varphi(\boldsymbol{x},0) \, d \boldsymbol{x} - \int_{0}^\infty \int_{\mathbb{R}^n} [u(\boldsymbol{x},t) \varphi_t(\boldsymbol{x},t) + \boldsymbol{f}(u(\boldsymbol{x},t)) \cdot \nabla \varphi(\boldsymbol{x},t)] \, d\boldsymbol{x}dt \end{aligned}\] by equation \(\eqref{Q5:3}\). Therefore \(u\) is a weak solution, as required.
Let \(u \in L^\infty(\mathbb{R}^n \times [0,\infty)) \cap C^1(\mathbb{R}^n \times [0,\infty))\) satisfy \[\begin{equation} \label{Q5:4} \int_{0}^\infty \int_{\mathbb{R}^n} [u(\boldsymbol{x},t) \varphi_t(\boldsymbol{x},t) + \boldsymbol{f}(u(\boldsymbol{x},t)) \cdot \nabla \varphi(\boldsymbol{x},t)] \, d \boldsymbol{x}dt + \int_{\mathbb{R}^n} u_0(\boldsymbol{x}) \varphi(\boldsymbol{x},0) \, d \boldsymbol{x}= 0 \end{equation}\] for all \(\varphi \in C_c^\infty(\mathbb{R}^n \times [0,\infty))\). Choose a test function \(\varphi \in C_c^\infty(\mathbb{R}^n \times [0,\infty))\) such that \(\mathrm{supp}(\varphi) \subset \mathbb{R}^n \times (0,\infty)\). In particular, \(\varphi(\boldsymbol{x},0)=0\). By equation \(\eqref{Q5:4}\), \(u\) satisfies \[\int_{0}^\infty \int_{\mathbb{R}^n} [u(\boldsymbol{x},t) \varphi_t(\boldsymbol{x},t) + \boldsymbol{f}(u(\boldsymbol{x},t)) \cdot \nabla \varphi(\boldsymbol{x},t)] \, d \boldsymbol{x}dt = 0.\] Since \(u\) is continuously differentiable, we can rewrite this equation using the integration by parts formula from above to obtain \[\int_0^\infty \int_{\mathbb{R}^n} [u_t(\boldsymbol{x},t) + \mathrm{div} \boldsymbol{f}(u(\boldsymbol{x},t))] \varphi(\boldsymbol{x},t) \, d \boldsymbol{x}dt = 0.\] But this holds for all \(\varphi \in C_c^\infty(\mathbb{R}^n \times [0,\infty))\) and so, by the Fundamental Lemma of the Calculus of Variations, the integrand must vanish: \[\begin{equation} \label{Q5:5} u_t(\boldsymbol{x},t) + \mathrm{div} \boldsymbol{f}(u(\boldsymbol{x},t)) = 0, \quad \forall \; (\boldsymbol{x},t) \in \mathbb{R}^n \times (0,\infty). \end{equation}\] Therefore \(u\) satisfies \(\eqref{Q5:2}\). It remains to show that \(u\) satisfies \(\eqref{Q5:3}\). Now take any test function \(\varphi \in C_c^\infty(\mathbb{R}^n \times [0,\infty))\). Integrating by parts in \(\eqref{Q5:4}\) gives \[\int_0^\infty \int_{\mathbb{R}^n} \underbrace{[u_t(\boldsymbol{x},t) + \mathrm{div} \boldsymbol{f}(u(\boldsymbol{x},t))]}_{=0 \textrm{ by } \eqref{Q5:5}} \varphi(\boldsymbol{x},t) \, d \boldsymbol{x}dt = \int_{\mathbb{R}^n} [u_0(\boldsymbol{x})-u(\boldsymbol{x},0)] \varphi(\boldsymbol{x},0) \, d \boldsymbol{x}.\] Therefore \[\int_{\mathbb{R}^n} [u_0(\boldsymbol{x})-u(\boldsymbol{x},0)] \varphi(\boldsymbol{x},0) \, d \boldsymbol{x}= 0\] for all \(\varphi \in C_c^\infty(\mathbb{R}^n \times [0,\infty))\). Applying the Fundamental Lemma of the Calculus of Variations again gives \(u(\boldsymbol{x},0)=u_0(\boldsymbol{x})\), as required.
The Rankine-Hugoniot condition. Consider Burger’s equation with initial data \[u_0(x) = \left\{ \begin{array}{cl} 2 & \textrm{if } x<0, \\ 1 & \textrm{if } 0<x<1, \\ 0 & \textrm{if } x>1. \end{array} \right.\]
Since \(u_0\) is discontinuous at \(x=0\) with a left limit of \(2\) and a right limit of \(1\), we seek a shock \(x = \sigma_1(t)\) through \((x,t)=(0,0)\) with \(u=2\) on the left of the shock and \(u=1\) on the right of the shock. The Rankine-Hugoniot condition with \(f(u)=\tfrac12 u^2\), \(u_l=2\), \(u_r=1\) gives \[\dot{\sigma}_1 = \frac{[[f(u)]]}{[[u]]} = \frac{\tfrac12 2^2 - \tfrac12 1^2}{2-1} = \frac 32.\] The conditions \(\dot{\sigma}_1(t)=3/2\) and \(\sigma_1(0)=0\) imply that \[\boxed{\sigma_1(t) = \frac{3}{2} t}\] Since \(u_0\) is discontinuous at \(x=1\) with a left limit of \(1\) and a right limit of \(0\), we seek a shock \(x = \sigma_2(t)\) through \((x,t)=(1,0)\) with \(u=1\) on the left of the shock and \(u=0\) on the right of the shock. The Rankine-Hugoniot condition with \(f(u)=\tfrac12 u^2\), \(u_l=1\), \(u_r=0\) gives \[\dot{\sigma}_2 = \frac{[[f(u)]]}{[[u]]} = \frac{\tfrac12 1^2 - \tfrac12 0^2}{1-0} = \frac 12.\] The conditions \(\dot{\sigma}_2(t)=1/2\) and \(\sigma_2(0)=1\) imply that \[\boxed{\sigma_2(t) = \frac{1}{2} t + 1}\] The shocks are straight lines and they intersect when \[\sigma_1(t) = \sigma_2(t) \quad \Longleftrightarrow \quad \frac 32 t = \frac 12 t + 1 \quad \Longleftrightarrow \quad t = 1.\] Therefore the intersection point is \((x,t)=(\sigma_1(1),1) = (3/2,1)\), as desired.
We have found the following weak solution for \(0<t<1\): \[u(x,t) = \left\{ \begin{array}{cl} 2 & \textrm{if } x < \sigma_1(t), \\ 1 & \textrm{if } \sigma_1(t) < x < \sigma_2(t), \\ 0 & \textrm{if } x > \sigma_2(t), \end{array} \right. \quad \Longleftrightarrow \quad \boxed{ u(x,t) = \left\{ \begin{array}{cl} 2 & \textrm{if } x < \tfrac32 t, \\ 1 & \textrm{if } \tfrac 32 t< x < \tfrac12 t + 1, \\ 0 & \textrm{if } x > \tfrac12 t + 1. \end{array} \right.}\]
We seek a shock \(x = \sigma_3(t)\) through \((x,t)=(3/2,1)\) with \(u=2\) on the left of the shock and \(u=0\) on the right of the shock. The Rankine-Hugoniot condition with \(f(u)=\tfrac12 u^2\), \(u_l=2\), \(u_r=0\) gives \[\dot{\sigma}_3 = \frac{[[f(u)]]}{[[u]]} = \frac{\tfrac12 2^2 - \tfrac12 0^2}{2-0} = 1.\] The conditions \(\dot{\sigma}_3(t)=1\) and \(\sigma_3(1)=3/2\) imply that \[\boxed{\sigma_3(t) = t + \tfrac12}\] We have found the following weak solution for all \(t \ge 0\): \[\boxed{ u(x,t) = \left\{ \begin{array}{cl} 2 & \textrm{if } 0 < t <1, \, x < \tfrac32 t, \\ 1 & \textrm{if } 0 < t <1, \, \tfrac 32 t< x < \tfrac12 t + 1, \phantom{\bigg|} \\ 0 & \textrm{if } 0 < t <1, \, x > \tfrac12 t + 1. \\ 2 & \textrm{if } t >1, \, x < t + \tfrac12, \phantom{\bigg|} \\ 0 & \textrm{if } t >1, \, x > t + \tfrac12. \end{array} \right.}\] The characteristic through \((x,t)=(s,0)\) is the line with equation \[x = s + c(u_0(s))t = \left\{ \begin{array}{cl} s + 2t & \textrm{if } s < 0, \\ s + t & \textrm{if } 0 < s < 1, \\ s & \textrm{if } s > 1. \end{array} \right.\] The characteristics and the shocks are plotted in Figure 2. In the blue region \(u=2\), in the magenta region \(u=1\), and in the green region \(u=0\). The red dashed lines are the shocks.
Since we are in the case of a uniformly strictly convex flux function, i.e. \(f(s)=\frac12s^2\), from the lectures we know that a shock satisfies Lax’s entropy condition if and only if \(u_r<u_l\) (where \(u_l\) and \(u_r\) are the left and right limits of the solution at the shock curves). As we see this from the definition of the weak solution, at every shock we have this condition satisfied. So, since all shocks satisfy Lax’s entropy condition, so does the constructed weak solution.
For the entropy solution, it remains to only check whether the constructed solution is uniformly bounded and satisfies the following one sided jump condition property. There exists \(C>0\) such that \[u(x+z,t)- u(x,t) \le C(1+1/t)z,\ \ \forall (x,t)\in\mathbb{R}\times(0,+\infty),\ \ \forall\ z>0.\] Since for each fixed time slice \(t\), we have that \(x\mapsto u(x,t)\) is non-increasing, we certainly have that \[u(x+z,t)- u(x,t) \le 0,\ \ \forall (x,t)\in\mathbb{R}\times(0,+\infty),\ \ \forall\ z>0.\] Also, clearly, the solution is uniformly bounded. Therefore, this solution is certainly the unique entropy solution.
The Lax entropy condition. We need to find \(a,b,c,d>0\) so that the conservation law \(u_t+uu_x=0\) is satisfied piecewise, so that the Rankine-Hugoniot condition is satisfied across the shock \(x=-dt^2\), and so that the Lax entropy condition holds. Let \[u(x,t) = \left\{ \begin{array}{ll} -a (t + \sqrt{bx+ct^2}) & \textrm{if } x > -dt^2, \\ 0 & \textrm{if } x < -dt^2. \end{array} \right.\] Define \[r = r(x,t) = \sqrt{bx + ct^2}.\] Clearly \(u_t+uu_x=0\) is satisfied for \(x<-dt^2\). Now consider the case \(x>-dt^2\). Then \[\begin{aligned} u_t = -a - \frac{act}{r}, \qquad u_x = -\frac{ab}{2r}. \end{aligned}\] Let’s check when the conservation law is satisfied: \[\begin{aligned} u_t + u u_x = 0 \quad & \Longleftrightarrow \quad -a - \frac{act}{r} +\frac{a^2 b}{2r}(t+r) = 0 %\\ %& \Longleftrightarrow \quad %\frac{-2ar -2act + a^2b(t+r)}{2r} = 0 \\ & \Longleftrightarrow \quad \frac{-2ar -2act + a^2b(t+r)}{2r} = 0 \\ & \Longleftrightarrow \quad -2r -2ct + ab(t+r) = 0 \\ & \Longleftrightarrow \quad t(ab-2c)+r(ab-2)=0. \end{aligned}\] Since this must hold for all \(x,t\) with \(x>-dt^2\), we have \(ab-2c=0\) and \(ab-2=0\). Therefore \[\boxed{ab=2} \qquad \boxed {c=1}\] Now we check when the Rankine-Hugoniot condition is satisfied. We have \[\sigma(t)=-dt^2, \quad f(u)=\frac12 u^2, \quad u_l = 0, \quad u_r = -a(t+\sqrt{b\sigma(t)+ct^2}) = -at (1+\sqrt{c-bd}).\] Therefore the Rankine-Hugoniot condition holds if and only if \[\begin{aligned} \dot{\sigma} = \frac{[[f(u)]]}{[[u]]} & \quad \Longleftrightarrow \quad -2dt = \frac{\frac12 u_l^2 - \frac12 u_r^2}{u_l - u_r} \\ %& \quad \Longleftrightarrow \quad -2dt = \frac{\frac12 u_l - \frac12 u_r}{u_l - u_r} %\\ & \quad \Longleftrightarrow \quad -2dt = \frac{1}{2} (u_l + u_r) \\ & \quad \Longleftrightarrow \quad - 4 dt = -at(1+\sqrt{c-bd}) \\ & \quad \Longleftrightarrow \quad \boxed{ 4d = a(1+\sqrt{c-bd})} \end{aligned}\] Finally, since \(c(u)=u\), the Lax entropy condition holds if and only if \[c(u_r) < \dot{\sigma} < c(u_l) \quad \Longleftrightarrow \quad -at(1+\sqrt{c-bd}) < -2dt < 0 \quad \Longleftrightarrow \quad -4dt < -2dt < 0.\] This is satisfied for all \(t>0\) if \[\boxed{d>0}\]
To check whether this solution is an entropy solution, we need to check whether is the solution uniformly bounded and is there \(C>0\) such that \[u(x+z,t)- u(x,t) \le C(1+1/t)z,\ \ \forall (x,t)\in\mathbb{R}\times(0,+\infty),\ \ \forall\ z>0.\] First, let us notice that the previously derived conditions on the constants imply that all these constants must be positive ones. Therefore, for each fixed time slice \(t\), we have that \(x\mapsto u(x,t)\) is non-increasing. Therefore, we have that \[u(x+z,t)- u(x,t) \le 0,\ \ \forall (x,t)\in\mathbb{R}\times(0,+\infty),\ \ \forall\ z>0.\] However, it is clear to see to this solution is not uniformly bounded, therefore by our definition it is not an entropy solution.
The Lax entropy condition again.
The flux is \(f(u)=\frac13 u^3\) and the shock has equation \(x=\sigma(t)\) with \(\sigma(t)=st\). The left and right limits of \(u\) at the shock are \(u_l=u_-\), \(u_r = u_+\). Therefore the Rankine-Hugoniot condition simplifies to \[\begin{aligned} \dot{\sigma} = \frac{[[f(u)]]}{[[u]]} \quad & \Longleftrightarrow \quad s = \frac{f(u_-)-f(u_+)}{u_- - u_+} \\ & \Longleftrightarrow \quad s = \frac{\frac13 u_-^3 - \frac13 u_+^3}{u_- - u_+} \\ & \Longleftrightarrow \quad s = \frac{(u_- - u_+)(u_+^2 + u_+ u_- + u_-^2)}{3(u_- - u_+)} \\ & \Longleftrightarrow \quad \boxed{s = \frac{u_+^2 + u_+ u_- + u_-^2}{3}} \end{aligned}\]
Since \(u_- = 1\), \[s = \frac{u_+^2 + u_+ + 1}{3}.\] We have \(c(u)=f'(u) = u^2\). The Lax entropy condition is \[c(u_+) < \dot{\sigma} < c(u_-) \quad \Longleftrightarrow \quad u_+^2 < s < 1 \quad \Longleftrightarrow \quad 3 u_+^2 < u_+^2 + u_+ + 1 < 3.\] Therefore we have to find \(u_+\) satisfying the quadratic inequalities \[u_+^2 + u_+ -2 < 0, \qquad 2u_+^2 - u_+ -1 < 0.\] Since \[u_+^2 + u_+ -2 = (u_+ +2)(u_+ -1), \qquad 2u_+^2 - u_+ -1 = (2u_+ + 1)(u_+ -1),\] then \(u_+\) must satisfy \(u_+ \in (-2,1)\) and \(u_+ \in (-1/2,1)\) and therefore \[\boxed{ u_+ \in (-1/2 ,1) }\]
For the entropy solution, we need to check whether the constructed solution in (ii) is uniformly bounded and satisfies the following one sided jump condition property: there exists \(C>0\) such that \[u(x+z,t)- u(x,t) \le C(1+1/t)z,\ \ \forall (x,t)\in\mathbb{R}\times(0,+\infty),\ \ \forall\ z>0.\] Since for each fixed time slice \(t\), we have that \(x\mapsto u(x,t)\) is non-increasing (since \(u_+<u_-\)), we certainly have that \[u(x+z,t)- u(x,t) \le 0,\ \ \forall (x,t)\in\mathbb{R}\times(0,+\infty),\ \ \forall\ z>0.\] Also, clearly the solution is uniformly bounded. Therefore, it is certainly the unique entropy solution.
Rarefaction waves and shocks.
The conservation law has \(c(u)=u^4\). Therefore the characteristics are \[x = s + c(u_0(s))t = \left\{ \begin{array}{cl} s & s<0, \\ s+t & 0 < s < 3, \\ s & s > 3. \end{array} \right.\] By sketching the characteristics (see Figure 3) we see that characteristics intersect immediately, at \((x,t)=(3,0)\). Therefore the conservation law does not have a classical solution for any \(t>0\).
We can take the flux to be \[\boxed{f(u)= \frac{1}{5} u^5}\] since \(f'(u)=u^4 = c(u)\). The Rankine-Hugoniot jump condition for a shock of the form \(x= \sigma(t)\) is \[\dot{\sigma} = \frac{[[f(u)]]}{[[u]]} \quad \Longleftrightarrow \quad \boxed{\dot{\sigma} = \frac{\frac{1}{5}(u_l^5 - u_r^5)}{u_l - u_r}}\]
From Figure 3 we see that there is a shock \(x = \sigma_1(t)\) with \(\sigma_1(0)=3\), \(u_l=1\), \(u_r=0\). The Rankine-Hugoniot condition gives \[\dot{\sigma}_1 = \frac{\frac{1}{5}(1^5 - 0^5)}{1 - 0} = \frac15.\] The conditions \(\dot{\sigma}_1=1/5\), \(\sigma_1(0)=3\) imply that \[\boxed{\sigma_1(t)=t/5 + 3}\] The characteristics and the shock are plotted in Figure 4. From this figure we see that \[u(x,t) = \left\{ \begin{array}{cl} 0 & \textrm{if } x<0, \\ 1 & \textrm{if } t < x < \sigma_1(t), \\ 0 & \textrm{if } x > \sigma_1(t). \end{array} \right.\]
The characteristics tell us nothing about the solution in the region \(0 < x < t\). We can fill the void \(0<x<t\) with lines of the form \(x/t = C\), \(C \in [0,1]\). These are shown in green in Figure 5.
We seek a rarefaction wave that is constant along these lines of the form \(u(x,t)=h(x/t)\) with \(h(0)=0\), \(h(1)=1\): \[u_t + u^4 u_x = 0 \quad \Longleftrightarrow \quad -\frac{x}{t^2} h' + h^4 \frac{1}{t} h' = 0 \quad \Longleftrightarrow \quad \frac{h'}{t} \left( h^4 - \frac{x}{t} \right) = 0.\] Therefore \(u(x,t)=h(x/t)=(x/t)^{1/4}\) for \(0 < x < t\). We have derived the following weak solution for \(0< t < 15/4\): \[u(x,t) = \left\{ \begin{array}{cl} 0 & \textrm{if } x<0, \\ (x/t)^{1/4} & \textrm{if } 0<x<t, \\ 1 & \textrm{if } t < x < \sigma_1(t), \\ 0 & \textrm{if } x > \sigma_1(t), \end{array} \right.\] where \(\sigma_1(t) = t/5 + 3\), as required.
For \(t\ge 15/4\) the expansion fan collides with the characteristics with \(s \ge 15/4\). The point of first intersection is \((x,t)=(15/4,15/4)\). (This can be found by finding the intersection of the lines \(x=t\) and \(x=\sigma_1(t)\).) Therefore we seek a shock \(x=\sigma_2(t)\) through \((x,t)=(15/4,15/4)\) with \(u(x,t)=(x/t)^{1/4}\) on the left of the shock and \(u=0\) on the right of the shock. Consequently \(u_l = (\sigma_2(t)/t)^{1/4}\), \(u_r=0\), and the Rankine-Hugoniot condition gives \[\dot{\sigma}_2 = \frac{[[f(u)]]}{[[u]]} = \frac{\frac15 (u_l^5 - u_r^5)}{u_l - u_r} = \frac15 u_l^4 = \frac{\sigma_2}{5t} \quad \Longleftrightarrow \quad \dot{\sigma}_2 - \frac{1}{5t} \sigma_2 = 0.\] We can solve this ODE using the integrating factor \[e^{- \int \frac{1}{5t} \, dt } = e^{-\frac 15 \ln t} = t^{-1/5}\] as follows: \[\dot{\sigma_2} - \frac{1}{5t} \sigma_2 = 0 \quad \Longleftrightarrow \quad t^{-1/5} \dot{\sigma}_2 - \frac{1}{5} t^{-6/5} \sigma_2 = 0 \quad \Longleftrightarrow \quad \frac{d}{dt} (t^{-1/5} \sigma_2) = 0.\] Integrating and using the condition \(\sigma_2(15/4) = 15/4\) yields \[t^{-1/5} \sigma_2(t) - \left( \tfrac{15}{4} \right)^{-1/5} \sigma_2 \left( \tfrac{15}{4} \right) = 0 %\left( \frac 32 \right)^{2/3} \quad \Longleftrightarrow \quad \boxed{\sigma_2(t) = \left( \tfrac {15}{4} \right)^{4/5} t^{1/5}}\] The shock is easier to sketch if we write it in the form \(t = (\frac{4}{15})^{4} x^5\). We have found the weak solution \[\boxed{ u(x,t) = \left\{ \begin{array}{cl} 0 & \textrm{if } 0<t<15/4, \; x<0, \\ (x/t)^{1/4} & \textrm{if } 0<t<15/4, \; 0<x<t, \\ 1 & \textrm{if } 0<t<15/4, \; t < x < \sigma_1(t), \\ 0 & \textrm{if } 0<t<15/4, \; x > \sigma_1(t), \\ 0 & \textrm{if } t>15/4, \; x<0, \\ (x/t)^{1/4} & \textrm{if } t>15/4, \; 0<x<\sigma_2(t), \\ 0 & \textrm{if } t>15/4, \; x > \sigma_2(t), \end{array} \right. }\] where \(\sigma_1(t) = t/5 + 3\), \(\sigma_2(t) = (\frac {15}{4})^{\frac 45} t^{\frac 15}\). This is illustrated in Figure 6.
Watching paint dry.
The conservation law has flux \(f(u)=\frac13 u^3\) and \(c(u)=f'(u)=u^2\). The characteristics are \[x = s + c(u_0(s))t = \left\{ \begin{array}{cl} s & s<0, \\ s+t & 0 < s < 1, \\ s & s > 1. \end{array} \right.\] By sketching the characteristics (see Figure 7) we see that characteristics intersect immediately, at \((x,t)=(1,0)\). Therefore we seek a shock \(x = \sigma(t)\) with \(\sigma(0)=1\), \(u_l=1\), \(u_r=0\). The Rankine-Hugoniot condition gives \[\dot{\sigma} = \frac{[[f(u)]]}{[[u]]} = \frac{\frac13 u_l^3 - \frac13 u_r^3}{u_l - u_r} = \frac13.\] The conditions \(\dot{\sigma}=1/3\), \(\sigma(0)=1\) imply that \(\sigma(t)=1+t/3\). The characteristics and the shock are plotted in Figure 8. From this figure we see that \[u(x,t) = \left\{ \begin{array}{cl} 0 & \textrm{if } x<0, \\ 1 & \textrm{if } t < x < \sigma(t), \\ 0 & \textrm{if } x > \sigma(t). \end{array} \right.\]
The characteristics tell us nothing about the solution in the region \(0 < x < t\). We can fill the void \(0<x<t\) with lines of the form \(x/t = C\), \(C \in [0,1]\). These are shown in green in Figure 9.
We seek a rarefaction wave that is constant along these lines of the form \(u(x,t)=h(x/t)\), \(h(0)=0\), \(h(1)=1\): \[u_t + u^2 u_x = 0 \quad \Longleftrightarrow \quad -\frac{x}{t^2} h' + h^2 \frac{1}{t} h' = 0 \quad \Longleftrightarrow \quad \frac{h'}{t} \left( h^2 - \frac{x}{t} \right) = 0.\] Therefore \(u(x,t)=h(x/t)=(x/t)^{1/2}\) for \(0 < x < t\). We have derived the following weak solution for \(0< t < 3/2\): \[u(x,t) = \left\{ \begin{array}{cl} 0 & \textrm{if } x<0, \\ (x/t)^{1/2} & \textrm{if } 0<x<t, \\ 1 & \textrm{if } t < x < \sigma(t), \\ 0 & \textrm{if } x > \sigma(t), \end{array} \right.\] where \(\sigma(t) = 1 + t/3\), as required.
For \(t\ge 3/2\) the expansion fan collides with the characteristics with \(s \ge 3/2\). The point of first intersection is \((x,t)=(3/2,3/2)\). (This can be found by finding the intersection of the lines \(x=t\) and \(x=\sigma(t)\).) Therefore we seek a shock \(x=\sigma_2(t)\) through \((x,t)=(3/2,3/2)\) with \(u(x,t)=(x/t)^{1/2}\) on the left of the shock and \(u=0\) on the right of the shock. Consequently \(u_l = (\sigma_2(t)/t)^{1/2}\), \(u_r=0\), and the Rankine-Hugoniot condition gives \[\dot{\sigma}_2 = \frac{[[f(u)]]}{[[u]]} = \frac{\frac13 u_l^3 - \frac13 u_r^3}{u_l - u_r} = \frac{\sigma_2}{3t} \quad \Longleftrightarrow \quad \dot{\sigma}_2 - \frac{1}{3t} \sigma_2 = 0.\] We can solve this ODE using the integrating factor \[e^{- \int \frac{1}{3t} \, dt } = e^{-\frac 13 \ln t} = t^{-1/3}\] as follows: \[\dot{\sigma_2} - \frac{1}{3t} \sigma_2 = 0 \quad \Longleftrightarrow \quad t^{-1/3} \dot{\sigma}_2 - \frac{1}{3} t^{-4/3} \sigma_2 = 0 \quad \Longleftrightarrow \quad \frac{d}{dt} (t^{-1/3} \sigma_2) = 0.\] Integrating and using the condition \(\sigma_2(3/2) = 3/2\) yields \[t^{-1/3} \sigma_2(t) - \left( \tfrac 32 \right)^{-1/3} \sigma_2 \left( \tfrac 32 \right) = 0 %\left( \frac 32 \right)^{2/3} \quad \Longleftrightarrow \quad \sigma_2(t) = \left( \tfrac 32 \right)^{2/3} t^{1/3}.\] The shock is easier to sketch if we write it in the form \(t = \frac 49 x^3\). We have found the weak solution \[\boxed{ u(x,t) = \left\{ \begin{array}{cl} 0 & \textrm{if } 0<t<3/2, \; x<0, \\ (x/t)^{1/2} & \textrm{if } 0<t<3/2, \; 0<x<t, \\ 1 & \textrm{if } 0<t<3/2, \; t < x < \sigma(t), \\ 0 & \textrm{if } 0<t<3/2, \; x > \sigma(t), \\ 0 & \textrm{if } t>3/2, \; x<0, \\ (x/t)^{1/2} & \textrm{if } t>3/2, \; 0<x<\sigma_2(t), \\ 0 & \textrm{if } t>3/2, \; x > \sigma_2(t), \end{array} \right. }\] where \(\sigma(t) = 1 + t/3\), \(\sigma_2(t) = (\frac 32)^{\frac 23} t^{\frac 13}\). This is illustrated in Figure 10.
Traffic flow.
Let \(f(\rho)=\rho v(\rho) = v_m \rho(1-\rho)\). Then \[f'(\rho) = v_m (1-\rho)-v_m \rho = v_m (1-2 \rho)\] and therefore \[\rho_t + f(\rho)_x = 0 \quad \Longleftrightarrow \quad \rho_t + v_m (1-2 \rho)\rho_x = 0\] as required.
Let \(c(\rho) = f'(\rho) = v_m(1-2 \rho)\). The characteristics for the conservation law are \[\rho = s + c(\rho_0(s))t = \left\{ \begin{array}{cl} s + \frac{v_m}{2} t & s<0, \\ s + \frac{v_m}{3} t & 0 < s < 1, \\ s + \frac{v_m}{2} t & s > 1. \end{array} \right.\] For \(s<0\), the characteristics have slope \(1/c(\rho_0(s)) = 2/ v_m\). The characteristics for \(0<s<1\) are steeper; they have slope \(1/c(\rho_0(s)) = 3/ v_m\). Therefore a shock forms at \((x,t)=(0,0)\). We seek a shock \(x = \sigma_1(t)\) through \((x,t)=(0,0)\) with \(\rho=\rho_l = 1/4\) on the left of the shock and \(\rho=\rho_r=1/3\) on the right of the shock. The Rankine-Hugoniot condition implies that \[\begin{aligned} \dot{\sigma_1}(t) & = \frac{[[f(\rho)]]}{[[\rho]]} \\ & = \frac{v_m \rho_l(1-p_l) - v_m \rho_r(1-\rho_r)}{\rho_l - \rho_r} \\ & = \frac{v_m (\rho_l - \rho_r)(1 - \rho_l - \rho_r)}{\rho_l - \rho_r} \\ & = v_m (1- \rho_l - \rho_r) \\ & = \frac{5}{12} v_m. \end{aligned}\] Therefore \(\sigma_1(t) = \frac{5}{12} v_m t\). The characteristics and the shock are shown in Figure 11.
The characteristics tell us nothing about the solution in the void \(1 + \frac{v_m}{3} t < x < 1+ \frac{v_m}{2} t\). We can fill in this void with lines of the form \((x-1)/t = C\), \(C \in [v_m/3,v_m/2]\). These are shown in green in Figure 12.
We seek a rarefaction wave that is constant along these lines of the form \(\rho(x,t)=h((x-1)/t)\) with \(h(v_m/3)=1/3\), \(h(v_m/2)=1/4\): \[\begin{aligned} \rho_t + v_m (1-2 \rho) \rho_x = 0 \quad & \Longleftrightarrow \quad -\frac{(x-1)}{t^2} h' + v_m (1-2h) \frac{1}{t} h' = 0 \\ & \Longleftrightarrow \quad \frac{h'}{t} \left( - \frac{(x-1)}{t} + v_m - 2 v_m h \right) = 0 \\ & \Longleftrightarrow \quad h \left( \frac{x-1}{t} \right) = \frac 12 - \frac{x-1}{2 v_m t} = \frac{v_m t - x +1}{2 v_m t}. \end{aligned}\] (As a consistency check, note that \(h(y) = \frac{1}{2} - \frac{y}{2 v_m}\) satisfies \(h(v_m/3)=1/3\), \(h(v_m/2)=1/4\) as desired.) Therefore \[\rho(x,t) = \frac{v_m t - x +1}{2 v_m t}\] for \(1 + \frac{v_m}{3} t < x < 1+ \frac{v_m}{2} t\), for sufficiently small \(t\). After some point in time the rarefaction wave collides with the characteristics with \(s<0\). See Figure 13.
This collision first occurs when \[\sigma_1(t) = 1+ \frac{v_m}{3} t \quad \Longleftrightarrow \quad t = \frac{12}{v_m}.\] The corresponding value of \(x\) is \[x = \sigma_1 \left( \tfrac{12}{v_m} \right) = 5.\] Now a second shock \(x=\sigma_2(t)\) forms with \(\sigma_2(\tfrac{12}{v_m})=5\), \(\rho = 1/4\) on the left of the shock, and \(\rho = (v_m t - x +1)/(2 v_m t)\) on the right of the shock. Therefore \(\rho_l = 1/4\), \(\rho_r = (v_m t - \sigma_2(t) +1)/(2 v_m t)\), and the Rankine-Hugoniot condition gives \[\dot{\sigma}_2(t) = v_m(1 - \rho_l - \rho_r) = \frac{v_m}{4} + \frac{\sigma_2(t)-1}{2t}.\] Multiplying this linear ODE by the integrating factor \[\exp \left( - \int \frac{1}{2t} \, dt \right) = t^{-1/2}\] gives \[\frac{d}{dt} (t^{-1/2} \sigma_2(t)) = \frac{v_m}{4} t^{-1/2} - \frac12 t^{-3/2}.\] Now integrate and use the initial condition \(\sigma_2(\tfrac{12}{v_m})=5\) to obtain \[\begin{aligned} t^{-1/2} \sigma_2(t) - \left( \tfrac{12}{v_m} \right)^{-1/2} \sigma_2 \left( \tfrac{12}{v_m} \right) & = \int_{12/v_m}^t \left( \frac{v_m}{4} s^{-1/2} - \frac12 s^{-3/2} \right) \, ds \\ & = \frac{v_m}{2} \left[ t^{1/2} - \left(\tfrac{12}{v_m}\right)^{1/2} \right] + t^{-1/2} - \left( \tfrac{12}{v_m} \right)^{-1/2}. \end{aligned}\] Therefore \[\sigma_2(t) = \frac{v_m t}{2} + 1 + \left( 4 \left( \tfrac{12}{v_m} \right)^{-1/2}- \frac{v_m}{2}\left( \tfrac{12}{v_m} \right)^{1/2} \right) t^{1/2} = \frac{v_m t}{2} + 1 - \left( \frac{v_m t}{3} \right)^{1/2}.\] We have arrived at the following weak solution: \[\boxed{ \rho(x,t) = \left\{ \begin{array}{cl} 1/4 & \textrm{if } 0<t<\frac{12}{v_m}, \; x<\sigma_1(t), \phantom{\Big|} \\ 1/3 & \textrm{if } 0<t<\frac{12}{v_m}, \; \sigma_1(t)<x<1+ \frac{v_m t}{3},\phantom{\Big|} \\ \displaystyle \frac{v_m t - x +1}{2 v_m t} & \textrm{if } 0<t<\frac{12}{v_m}, \; 1+ \frac{v_m t}{3} < x < 1+ \frac{v_m t}{2},\phantom{\Big|} \\ 1/4 & \textrm{if } 0<t<\frac{12}{v_m}, \; x > 1+ \frac{v_m t}{2},\phantom{\Big|} \\ 1/4 & \textrm{if } t>\frac{12}{v_m}, \; x<\sigma_2(t),\phantom{\Big|} \\ \displaystyle \frac{v_m t - x +1}{2 v_m t} & \textrm{if } t>\frac{12}{v_m}, \; \sigma_2(t)<x<1+ \frac{v_m t}{2},\phantom{\Big|} \\ 1/4 & \textrm{if } t>\frac{12}{v_m}, \; x > 1+ \frac{v_m t}{2},\phantom{\Big|} \end{array} \right. }\] where \[\boxed{ \sigma_1(t) = \frac{5 v_m t}{12}, \qquad \sigma_2(t) = \frac{v_m t}{2} + 1 - \left( \frac{v_m t}{3} \right)^{1/2} }\]
The solution \(\rho(x,t)\) as a function of \(x\) is illustrated in Figure 14 for times \(t=0\), \(6/v_m\), \(24/v_m\). The effect of the initial block of slower moving traffic never disappears completely and the shock exists for all time. Its effect, however, diminishes over time and the size of the jump discontinuity tends to zero. Observe that the width of the region where \(\rho > 1/4\) increases with time.
Strong solutions. The proof is somewhat similar to the derivation of the Rankine-Hugoniot condtion (or rather to the proof of the converse of the Rankine-Hugoniot condition). Let \(V = \mathbb{R} \times (0,\infty)\) and let \(\varphi \in C_c^\infty(V)\). Then \[\begin{aligned} & \int_V [u \varphi_t + f(u) \varphi_x] \, dx dt + \int_{-\infty}^\infty u_0(x) \varphi(x,0) \, dx \\ & = \int_{V_l} [u \varphi_t + f(u) \varphi_x] \, dx dt + \int_{V_r} [u \varphi_t + f(u) \varphi_x] \, dx dt + \int_{-\infty}^\infty u_0(x) \varphi(x,0) \, dx \\ & = \int_{V_l} (f(u),u) \cdot \nabla_{(x,t)} \varphi \, dx dt + \int_{V_r} (f(u),u) \cdot \nabla_{(x,t)} \varphi \, dx dt + \int_{-\infty}^\infty u_0(x) \varphi(x,0) \, dx \\ & = \int_{\partial V_l} (f(u),u) \varphi \cdot \boldsymbol{n}_l \, dL - \int_{V_l} \mathrm{div}_{(x,t)} (f(u),u) \varphi \, dx dt +\int_{\partial V_r} (f(u),u) \varphi \cdot \boldsymbol{n}_r \, dL - \int_{V_r} \mathrm{div}_{(x,t)} (f(u),u) \varphi \, dx dt \\ & \phantom{=} \; \; + \int_{-\infty}^\infty u_0(x) \varphi(x,0) \, dx \end{aligned}\] where \(\boldsymbol{n}_l\) and \(\boldsymbol{n}_r\) are the unit outward-pointing normal vectors to \(\partial V_l\), \(\partial V_r\). Observe that \[\partial V_l = \{ (x,0) : x \le g(0) \} \, \cup \, \gamma, \quad \partial V_r = \{ (x,0) : x \ge g(0) \} \, \cup \, \gamma.\] Moreover \(\boldsymbol{n}_r = - \boldsymbol{n}_l \textrm{ on } \gamma\), \(\boldsymbol{n}_l = -(0,1)\) on \(\partial V_l \setminus \gamma\), \(\boldsymbol{n}_r = -(0,1)\) on \(\partial V_r \setminus \gamma\). Continuing the calculation gives \[\begin{aligned} & \int_V [u \varphi_t + f(u) \varphi_x] \, dx dt + \int_{-\infty}^\infty u_0(x) \varphi(x,0) \, dx \\ & = - \int_{-\infty}^{g(0)} (f(u),u) \varphi \big|_{t=0} \cdot (0,1) \, dx + \int_{\gamma} (f(u),u) \varphi \cdot \boldsymbol{n}_l \, dL - \int_{V_l} (\underbrace{u_t + f(u)_x}_{=0} ) \varphi \, dx dt \\ & \phantom{=} \; \; - \int_{g(0)}^\infty (f(u),u) \varphi \big|_{t=0} \cdot (0,1) \, dx + \int_{\gamma} (f(u),u) \varphi \cdot \boldsymbol{n}_r \, dL - \int_{V_r} (\underbrace{u_t + f(u)_x}_{=0} ) \varphi \, dx dt + \int_{-\infty}^\infty u_0(x) \varphi(x,0) \, dx \\ & = - \int_{-\infty}^{g(0)} u(x,0) \varphi(x,0) \, dx + \int_{\gamma} (f(u),u) \varphi \cdot \boldsymbol{n}_l \, dL- \int_{V_l} (\underbrace{u_t + f(u)_x}_{=0} ) \varphi \, dx dt \\ & \phantom{=} \; \; - \int_{g(0)}^\infty u(x,0) \varphi(x,0) \, dx + \int_{\gamma} (f(u),u) \varphi \cdot \underbrace{\boldsymbol{n}_r}_{=-\boldsymbol{n}_l} \, dL- \int_{V_r} (\underbrace{u_t + f(u)_x}_{=0} ) \varphi \, dx dt + \int_{-\infty}^\infty u_0(x) \varphi(x,0) \, dx \\ & = \int_{\infty}^\infty (\underbrace{u_0(x) - u(x,0)}_{=0}) \varphi(x,0) \, dx \\ & = 0. \end{aligned}\] This holds for all \(\varphi \in C_c^\infty(V)\). Therefore \(u\) is a weak solution of the conservation law, as required.