Partial Differential Equations III/IV

[6pt] Exercise Sheet 3: Solutions

1. Existence of classical solutions. Consider the conservation law

   $u_t+u^3{u}_x=0$ $\text{for }(x,t)\in ℝ\times (0,\mathrm{\infty }),$

   $u(x,0)=u_0(x)$ $\text{for }x\in ℝ,$

   where

   We have $c(u)=u^3$ and

   Note that $c(u_0(s))$ is decreasing for $s\in (0,1)$. By Theorem 3.5, the conservation law has a classical solution $u:ℝ\times [0,t_{\mathrm{c}})\to ℝ$ where

   $$t_{\mathrm{c}}=\underset{s\in (0,1)}{inf}\frac{-1}{c{(u_0(s))}_s}=\underset{s\in (0,1)}{inf}\frac{1}{12s{(1-s^2)}^5}.$$

   Note that $1/(12s{(1-s^2)}^5)\to \mathrm{\infty }$ as $s\to 0,1$. Therefore its minimum in $(0,1)$ is attained at a critical point. We compute

   ${\displaystyle \frac{d}{ds}}{\displaystyle \frac{1}{12s{(1-s^2)}^5}}$ $={\displaystyle \frac{d}{ds}}{(12s{(1-s^2)}^5)}^{-1}$

   $=-{(12s{(1-s^2)}^5)}^{-2}(12{(1-s^2)}^5-120s^2{(1-s^2)}^4)$

   $={\displaystyle \frac{-12{(1-s^2)}^4(1-11s^2)}{{(12s{(1-s^2)}^5)}^2}}.$

   If follows that, for $s\in (0,1)$,

   $$\frac{d}{ds}\frac{1}{12s{(1-s^2)}^5}=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}1-11s^2=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}{s}^2=\frac{1}{11}.$$

   Therefore the only critical point of $1/(12s{(1-s^2)}^5)$ in $(0,1)$ is $s=1/\sqrt{11}$, and this is its global minimum point. Its minimum value is

   $$t_{\mathrm{c}}={\frac{1}{12s{(1-s^2)}^5}|}_{s=1/\sqrt{11}}=\frac{\sqrt{11}}{12{(10/11)}^5}=\boxed{\frac{{11}^5\sqrt{11}}{{10}^{5}12}\approx 0.445}$$

2. Breakdown of classical solutions. Consider the conservation law

   $u_t+u{u}_x=0$ $\text{for }(x,t)\in ℝ\times (0,\mathrm{\infty }),$

   $u(x,0)={\displaystyle \frac{1}{1+x^2}}$ $\text{for }x\in ℝ.$

   • We have $c(u)=u$ and $c(u_0(s))=u_0(s)={(1+s^2)}^{-1}$. Observe that 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

     $$t_{\mathrm{c}}=\underset{s\in >0}{inf}\frac{-1}{c{(u_0(s))}_s}=\underset{s>0}{inf}\frac{{(1+s^2)}^2}{2s},$$

     where

     $$x_c(t)=s_c+tc(u_0(s_c)).$$

     Let us find now $s_c$. Note that ${(1+s^2)}^2/(2s)\to \mathrm{\infty }$ as $s\to 0,\mathrm{\infty }$. Therefore its minimum in $(0,\mathrm{\infty })$ is attained at a critical point. We compute

     $$\frac{d}{ds}\frac{{(1+s^2)}^2}{2s}=\frac{8s^2(1+s^2)-2{(1+s^2)}^2}{4s^2}=\frac{(1+s^2)(6s^2-2)}{4s^2}.$$

     Therefore, for $s>0$,

     $$\frac{d}{ds}\frac{{(1+s^2)}^2}{2s}=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}6s^2-2=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}{s}^2=\frac{1}{3}.$$

     It follows that the only critical point of ${(1+s^2)}^2/(2s)$ in $(0,\mathrm{\infty })$ is $s_c=1/\sqrt{3}\approx 0.5774$, and this is its global minimum point we were looking for. Its minimum value is

     $$t_{\mathrm{c}}={\frac{{(1+s^2)}^2}{2s}|}_{s=1/\sqrt{3}}=\frac{8\sqrt{3}}{9}=\boxed{\frac{8}{3\sqrt{3}}\approx 1.5396}$$

     as required. By the same theorem, we also know that $u_t(x_c(t),t)$ blows up at $t_c$, provided $c(u_0(s_c))\ne 0$. By computing

     $$c(u_0(s_c))=\frac{3}{4}\ne 0,$$

     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

     $$x=s+c(u_0(s))t=s+\frac{t}{1+s^2}\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}t=(1+s^2)(x-s).$$

     The slope $1+s^2$ is decreasing for and increasing for $s>0$. Therefore the characteristics diverge for (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 $t_c$.

     

     Figure 1: Characteristics for Q2.

   • 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

     $$\boxed{u(x,t)=\frac{1}{1+{(x-u(x,t)t)}^2}}$$

3. 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

   $$t_{\mathrm{c}}=\frac{-1}{c^{\prime }(u_0(s_{\mathrm{c}}))u_0^{\prime }(s_{\mathrm{c}})}.$$

   We have

   $$h_s(s,t)=1+c{(u_0(s))}_{s}t=1+c^{\prime }(u_0(s))u_0^{\prime }(s)t.$$

   Therefore

   because $t>t_{\mathrm{c}}$. Since $h(s_{\mathrm{c}},t)=x$ and , then there exists $s_1>s_{\mathrm{c}}$ such that . But $h(s,t)\to \mathrm{\infty }$ as $s\to \mathrm{\infty }$. 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 $ℝ\ni s\mapsto c(u_0(s))$ is Lipschitz continuous. Actually, this is also Lipschitz continuous when one restricts the domain to , so define

   $$L_I:=\underset{s\in I}{sup}|c^{\prime }(u_0(s))u_0^{\prime }(s)|.$$

   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.

   $$t_c:=\underset{s\in I}{inf}-\frac{1}{c^{\prime }(u_0(s))u_0^{\prime }(s)}=\frac{1}{L_I},$$

   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 . In this case, we can write again

   $$|s_1-s_2|=t|c(u_0(s_2))-c(u_0(s_1))|=t|c^{\prime }(u_0(s_0))u_0^{\prime }(s_0)||s_1-s_2|,$$

   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 $c^{\prime }(u_0(s_0))u_0^{\prime }(s_0)\ge 0$. In this case if

   $$0\le c^{\prime }(u_0(s_0))u_0^{\prime }(s_0)\le L_I,$$

   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

   $$c^{\prime }(u_0(s_0))u_0^{\prime }(s_0)>L_I.$$

   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 or at least $c^{\prime }(u_0(s_0))u_0^{\prime }(s_0)\le L_I$.

4. The Fundamental Lemma of the Calculus of Variations. Let $g:\mathrm{\Omega }\to ℝ$ be a continuous function on an open set $\mathrm{\Omega }\subseteq {ℝ}^n$. Suppose that

   $${\int }_{\mathrm{\Omega }}g(𝒙)\psi (𝒙)𝑑𝒙=0\mathit{\hspace{1em}}\text{for all }\psi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega }).$$ (1)

   We show that $g=0$. Assume for contradiction that $g\ne 0$. Without loss of generality, assume that there exists a point $𝒑\in \mathrm{\Omega }$ such that $g(𝒑)>0$. Since $g$ is continuous, then there exists an open set $U\subset \mathrm{\Omega }$ containing $𝒑$ such that $g(𝒙)>0$ for all $𝒙\in U$. Choose $\epsilon >0$ such that $B_{\epsilon }(𝒑)\subset U$, where $B_{\epsilon }(𝒑)$ is the open ball of radius $\epsilon$ centred at $𝒑$. Choose any nonzero $\phi \in C_c^{\mathrm{\infty }}(B_{\epsilon }(𝒑))$ such $\phi \ge 0$, i.e., $\phi (𝒙)\ge 0$ for all $𝒙\in B_{\epsilon }(𝒑)$. For example, we could take

   Then

   $${\int }_{\mathrm{\Omega }}g(𝒙)\phi (𝒙)𝑑𝒙={\int }_{B_{\epsilon }(𝒑)}g(𝒙)\phi (𝒙)𝑑𝒙>0,$$

   which contradicts equation (1). Therefore $g=0$, as required.

5. 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 $f^{\prime }(u)=c(u)=\ln |u|+1$. Note that $f$ is only unique up to a constant. We can choose

   $f(u)$ $={\displaystyle {\int }_0^u}(\ln |s|+1)𝑑s={\displaystyle {\int }_0^u}\left({\displaystyle \frac{ds}{ds}}\ln |s|+1\right)𝑑s$

   $={s\ln |s||}_0^u-{\displaystyle {\int }_0^u}s{\displaystyle \frac{d}{ds}}\ln |s|ds+{\displaystyle {\int }_0^u}1𝑑s=u\ln |u|-{\displaystyle {\int }_0^u}1𝑑s+{\displaystyle {\int }_0^u}1𝑑s$

   $=u\ln |u|.$

   Therefore $u\in L^{\mathrm{\infty }}(ℝ\times (0,\mathrm{\infty }))$ is a weak solution of the conservation law if

   $$\boxed{{\int }_{t=0}^{\mathrm{\infty }}{\int }_{x=-\mathrm{\infty }}^{\mathrm{\infty }}[u(x,t){\phi }_t(x,t)+u(x,t)\ln |u(x,t)|{\phi }_x(x,t)]𝑑x𝑑t+{\int }_{-\pi }^{\pi }(1+\cos x)\phi (x,0)𝑑x=0}$$

   for all $\phi \in C_c^{\mathrm{\infty }}(ℝ\times [0,\mathrm{\infty }))$.

6. 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 $u:{ℝ}^n\times [0,\mathrm{\infty })\to ℝ$, $𝒗:{ℝ}^n\times [0,\mathrm{\infty })\to {ℝ}^n$ be continuously differentiable and let $\phi \in C_c^{\mathrm{\infty }}({ℝ}^n\times [0,\mathrm{\infty }))$. Then

   $$\begin{array}{c}{\int }_0^{\mathrm{\infty }}{\int }_{{ℝ}^n}[u_t(𝒙,t)+\mathrm{div}𝒗(𝒙,t)]\phi (𝒙,t)𝑑𝒙𝑑t=\hfill \\ \hfill -{\int }_{{ℝ}^n}u(𝒙,0)\phi (𝒙,0)𝑑𝒙-{\int }_0^{\mathrm{\infty }}{\int }_{{ℝ}^n}[u(𝒙,t){\phi }_t(𝒙,t)+𝒗(𝒙,t)\cdot \nabla \phi (𝒙,t)]𝑑𝒙𝑑t.\end{array}$$

   • Let $u\in C^1({ℝ}^n\times [0,\mathrm{\infty }))$ be bounded and satisfy

     $u_t+\mathrm{div}𝒇(u)=0$ $\text{for }(𝒙,t)\in {ℝ}^n\times (0,\mathrm{\infty }),$ (2)

     $u(𝒙,0)=u_0(𝒙)$ $\text{for }𝒙\in {ℝ}^n.$ (3)

     Multiply equation (2) by $\phi \in C_c^{\mathrm{\infty }}({ℝ}^n\times [0,\mathrm{\infty }))$, integrate over ${ℝ}^n\times [0,\mathrm{\infty })$, and apply the integration by parts formula with $𝒗(𝒙,t)=𝒇(u(x,t))$ to obtain

     $0$ $={\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle {\int }_{{ℝ}^n}}[u_t(𝒙,t)+{\mathrm{div}}_{𝒙}𝒇(u(𝒙,t))]\phi (𝒙,t)𝑑𝒙𝑑t$

     $=-{\displaystyle {\int }_{{ℝ}^n}}u(𝒙,0)\phi (𝒙,0)𝑑𝒙-{\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle {\int }_{{ℝ}^n}}[u(𝒙,t){\phi }_t(𝒙,t)+𝒇(u(𝒙,t))\cdot \nabla \phi (𝒙,t)]𝑑𝒙𝑑t$

     $=-{\displaystyle {\int }_{{ℝ}^n}}{u}_0(𝒙)\phi (𝒙,0)𝑑𝒙-{\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle {\int }_{{ℝ}^n}}[u(𝒙,t){\phi }_t(𝒙,t)+𝒇(u(𝒙,t))\cdot \nabla \phi (𝒙,t)]𝑑𝒙𝑑t$

     by equation (3). Therefore $u$ is a weak solution, as required.

   • Let $u\in L^{\mathrm{\infty }}({ℝ}^n\times [0,\mathrm{\infty }))\cap C^1({ℝ}^n\times [0,\mathrm{\infty }))$ satisfy

     $${\int }_0^{\mathrm{\infty }}{\int }_{{ℝ}^n}[u(𝒙,t){\phi }_t(𝒙,t)+𝒇(u(𝒙,t))\cdot \nabla \phi (𝒙,t)]𝑑𝒙𝑑t+{\int }_{{ℝ}^n}{u}_0(𝒙)\phi (𝒙,0)𝑑𝒙=0$$ (4)

     for all $\phi \in C_c^{\mathrm{\infty }}({ℝ}^n\times [0,\mathrm{\infty }))$. Choose a test function $\phi \in C_c^{\mathrm{\infty }}({ℝ}^n\times [0,\mathrm{\infty }))$ such that $\mathrm{supp}(\phi )\subset {ℝ}^n\times (0,\mathrm{\infty })$. In particular, $\phi (𝒙,0)=0$. By equation (4), $u$ satisfies

     $${\int }_0^{\mathrm{\infty }}{\int }_{{ℝ}^n}[u(𝒙,t){\phi }_t(𝒙,t)+𝒇(u(𝒙,t))\cdot \nabla \phi (𝒙,t)]𝑑𝒙𝑑t=0.$$

     Since $u$ is continuously differentiable, we can rewrite this equation using the integration by parts formula from above to obtain

     $${\int }_0^{\mathrm{\infty }}{\int }_{{ℝ}^n}[u_t(𝒙,t)+\mathrm{div}𝒇(u(𝒙,t))]\phi (𝒙,t)𝑑𝒙𝑑t=0.$$

     But this holds for all $\phi \in C_c^{\mathrm{\infty }}({ℝ}^n\times [0,\mathrm{\infty }))$ and so, by the Fundamental Lemma of the Calculus of Variations, the integrand must vanish:

     $$u_t(𝒙,t)+\mathrm{div}𝒇(u(𝒙,t))=0,\forall (𝒙,t)\in {ℝ}^n\times (0,\mathrm{\infty }).$$ (5)

     Therefore $u$ satisfies (2). It remains to show that $u$ satisfies (3). Now take any test function $\phi \in C_c^{\mathrm{\infty }}({ℝ}^n\times [0,\mathrm{\infty }))$. Integrating by parts in (4) gives

     $${\int }_0^{\mathrm{\infty }}{\int }_{{ℝ}^n}\underset{=0\text{ by }(\mathit{\text{<a href="\#S0.E5" title="(5) ‣ item (ii) ‣ item 6" class="ltx\_ref ltx\_font\_italic" style="font-size:70\%;"><span class="ltx\_text ltx\_ref\_tag">5</span></a>}})}{\underbrace{[u_t(𝒙,t)+\mathrm{div}𝒇(u(𝒙,t))]}}\phi (𝒙,t)𝑑𝒙𝑑t={\int }_{{ℝ}^n}[u_0(𝒙)-u(𝒙,0)]\phi (𝒙,0)𝑑𝒙.$$

     Therefore

     $${\int }_{{ℝ}^n}[u_0(𝒙)-u(𝒙,0)]\phi (𝒙,0)𝑑𝒙=0$$

     for all $\phi \in C_c^{\mathrm{\infty }}({ℝ}^n\times [0,\mathrm{\infty }))$. Applying the Fundamental Lemma of the Calculus of Variations again gives $u(𝒙,0)=u_0(𝒙)$, as required.

7. The Rankine-Hugoniot condition. Consider Burger’s equation with initial data

   • 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 $f(u)=\frac{1}{2}{u}^2$, $u_l=2$, $u_r=1$ gives

     $${\dot{\sigma }}_1=\frac{[[f(u)]]}{[[u]]}=\frac{\frac{1}{2}{2}^2-\frac{1}{2}{1}^2}{2-1}=\frac{3}{2}.$$

     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 $f(u)=\frac{1}{2}{u}^2$, $u_l=1$, $u_r=0$ gives

     $${\dot{\sigma }}_2=\frac{[[f(u)]]}{[[u]]}=\frac{\frac{1}{2}{1}^2-\frac{1}{2}{0}^2}{1-0}=\frac{1}{2}.$$

     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

     $${\sigma }_1(t)={\sigma }_2(t)\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}\frac{3}{2}t=\frac{1}{2}t+1\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}t=1.$$

     Therefore the intersection point is $(x,t)=({\sigma }_1(1),1)=(3/2,1)$, as desired.

   • We have found the following weak solution for :

   • 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 $f(u)=\frac{1}{2}{u}^2$, $u_l=2$, $u_r=0$ gives

     $${\dot{\sigma }}_3=\frac{[[f(u)]]}{[[u]]}=\frac{\frac{1}{2}{2}^2-\frac{1}{2}{0}^2}{2-0}=1.$$

     The conditions ${\dot{\sigma }}_3(t)=1$ and ${\sigma }_3(1)=3/2$ imply that

     $$\boxed{{\sigma }_3(t)=t+\frac{1}{2}}$$

     We have found the following weak solution for all $t\ge 0$:

     The characteristic through $(x,t)=(s,0)$ is the line with equation

     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.

     

     Figure 2: Characteristics and shocks for Q7.

   • Since we are in the case of a uniformly strictly convex flux function, i.e. $f(s)=\frac{1}{2}{s}^2$, from the lectures we know that a shock satisfies Lax’s entropy condition if and only if (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

     $$u(x+z,t)-u(x,t)\le C(1+1/t)z,\forall (x,t)\in ℝ\times (0,+\mathrm{\infty }),\forall z>0.$$

     Since for each fixed time slice $t$, we have that $x\mapsto u(x,t)$ is non-increasing, we certainly have that

     $$u(x+z,t)-u(x,t)\le 0,\forall (x,t)\in ℝ\times (0,+\mathrm{\infty }),\forall z>0.$$

     Also, clearly, the solution is uniformly bounded. Therefore, this solution is certainly the unique entropy solution.

8. The Lax entropy condition. We need to find $a,b,c,d>0$ so that the conservation law $u_t+u{u}_x=0$ is satisfied piecewise, so that the Rankine-Hugoniot condition is satisfied across the shock $x=-d{t}^2$, and so that the Lax entropy condition holds. Let

   Define

   $$r=r(x,t)=\sqrt{bx+c{t}^2}.$$

   Clearly $u_t+u{u}_x=0$ is satisfied for . Now consider the case $x>-d{t}^2$. Then

   $u_t=-a-{\displaystyle \frac{act}{r}},u_x=-{\displaystyle \frac{ab}{2r}}.$

   Let’s check when the conservation law is satisfied:

   $u_t+u{u}_x=0$ $⟺\mathit{\hspace{1em}}-a-{\displaystyle \frac{act}{r}}+{\displaystyle \frac{a^{2}b}{2r}}(t+r)=0$

   $⟺\mathit{\hspace{1em}}{\displaystyle \frac{-2ar-2act+a^{2}b(t+r)}{2r}}=0$

   $⟺\mathit{\hspace{1em}}-2r-2ct+ab(t+r)=0$

   $⟺\mathit{\hspace{1em}}t(ab-2c)+r(ab-2)=0.$

   Since this must hold for all $x,t$ with $x>-d{t}^2$, we have $ab-2c=0$ and $ab-2=0$. Therefore

   $$\boxed{ab=2}\mathit{\hspace{1em}\hspace{1em}}\boxed{c=1}$$

   Now we check when the Rankine-Hugoniot condition is satisfied. We have

   $$\sigma (t)=-d{t}^2,f(u)=\frac{1}{2}{u}^2,u_l=0,u_r=-a(t+\sqrt{b\sigma (t)+c{t}^2})=-at(1+\sqrt{c-bd}).$$

   Therefore the Rankine-Hugoniot condition holds if and only if

   $\dot{\sigma }={\displaystyle \frac{[[f(u)]]}{[[u]]}}$ $\mathrm{\hspace{1em}}⟺\mathit{\hspace{1em}}-2dt={\displaystyle \frac{\frac{1}{2}{u}_l^2-\frac{1}{2}{u}_r^2}{u_l-u_r}}$

   $\mathrm{\hspace{1em}}⟺\mathit{\hspace{1em}}-2dt={\displaystyle \frac{1}{2}}(u_l+u_r)$

   $\mathrm{\hspace{1em}}⟺\mathit{\hspace{1em}}-4dt=-at(1+\sqrt{c-bd})$

   $\mathrm{\hspace{1em}}⟺\mathit{\hspace{1em}}\boxed{4d=a(1+\sqrt{c-bd})}$

   Finally, since $c(u)=u$, the Lax entropy condition holds if and only if

   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

   $$u(x+z,t)-u(x,t)\le C(1+1/t)z,\forall (x,t)\in ℝ\times (0,+\mathrm{\infty }),\forall z>0.$$

   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

   $$u(x+z,t)-u(x,t)\le 0,\forall (x,t)\in ℝ\times (0,+\mathrm{\infty }),\forall z>0.$$

   However, it is clear to see to this solution is not uniformly bounded, therefore by our definition it is not an entropy solution.

9. The Lax entropy condition again.

   • The flux is $f(u)=\frac{1}{3}{u}^3$ 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

     $\dot{\sigma }={\displaystyle \frac{[[f(u)]]}{[[u]]}}$ $⟺\mathit{\hspace{1em}}s={\displaystyle \frac{f(u_{-})-f(u_{+})}{u_{-}-u_{+}}}$

     $⟺\mathit{\hspace{1em}}s={\displaystyle \frac{\frac{1}{3}{u}_{-}^3-\frac{1}{3}{u}_{+}^3}{u_{-}-u_{+}}}$

     $⟺\mathit{\hspace{1em}}s={\displaystyle \frac{(u_{-}-u_{+})(u_{+}^2+u_{+}{u}_{-}+u_{-}^2)}{3(u_{-}-u_{+})}}$

     $⟺\mathit{\hspace{1em}}\boxed{s={\displaystyle \frac{u_{+}^2+u_{+}{u}_{-}+u_{-}^2}{3}}}$

   • Since $u_{-}=1$,

     $$s=\frac{u_{+}^2+u_{+}+1}{3}.$$

     We have $c(u)=f^{\prime }(u)=u^2$. The Lax entropy condition is

     Therefore we have to find $u_{+}$ satisfying the quadratic inequalities

     Since

     $$u_{+}^2+u_{+}-2=(u_{+}+2)(u_{+}-1),2u_{+}^2-u_{+}-1=(2u_{+}+1)(u_{+}-1),$$

     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

     $$u(x+z,t)-u(x,t)\le C(1+1/t)z,\forall (x,t)\in ℝ\times (0,+\mathrm{\infty }),\forall z>0.$$

     Since for each fixed time slice $t$, we have that $x\mapsto u(x,t)$ is non-increasing (since ), we certainly have that

     $$u(x+z,t)-u(x,t)\le 0,\forall (x,t)\in ℝ\times (0,+\mathrm{\infty }),\forall z>0.$$

     Also, clearly the solution is uniformly bounded. Therefore, it is certainly the unique entropy solution.

10. Rarefaction waves and shocks.

    • The conservation law has $c(u)=u^4$. Therefore the characteristics are

      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 $t>0$.

      

      Figure 3: Characteristics for Q10.

    • We can take the flux to be

      $$\boxed{f(u)=\frac{1}{5}{u}^5}$$

      since $f^{\prime }(u)=u^4=c(u)$. The Rankine-Hugoniot jump condition for a shock of the form $x=\sigma (t)$ is

      $$\dot{\sigma }=\frac{[[f(u)]]}{[[u]]}\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}\boxed{\dot{\sigma }=\frac{\frac{1}{5}(u_l^5-u_r^5)}{u_l-u_r}}$$

    • 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

      $${\dot{\sigma }}_1=\frac{\frac{1}{5}(1^5-0^5)}{1-0}=\frac{1}{5}.$$

      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

      

      Figure 4: Characteristics and the shock $x={\sigma }_1(t)$ for Q10.

      The characteristics tell us nothing about the solution in the region . We can fill the void with lines of the form $x/t=C$, $C\in [0,1]$. These are shown in green in Figure 5.

      

      Figure 5: Characteristics, the shock $x={\sigma }_1(t)$, and the expansion fan for Q10.

      We seek a rarefaction wave that is constant along these lines of the form $u(x,t)=h(x/t)$ with $h(0)=0$, $h(1)=1$:

      $$u_t+u^4{u}_x=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}-\frac{x}{t^2}{h}^{\prime }+h^4\frac{1}{t}{h}^{\prime }=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}\frac{h^{\prime }}{t}\left(h^4-\frac{x}{t}\right)=0.$$

      Therefore $u(x,t)=h(x/t)={(x/t)}^{1/4}$ for . We have derived the following weak solution for :

      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

      $${\dot{\sigma }}_2=\frac{[[f(u)]]}{[[u]]}=\frac{\frac{1}{5}(u_l^5-u_r^5)}{u_l-u_r}=\frac{1}{5}{u}_l^4=\frac{{\sigma }_2}{5t}\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}{\dot{\sigma }}_2-\frac{1}{5t}{\sigma }_2=0.$$

      We can solve this ODE using the integrating factor

      $$e^{-\int \frac{1}{5t}𝑑t}=e^{-\frac{1}{5}\ln t}=t^{-1/5}$$

      as follows:

      $$\dot{{\sigma }_2}-\frac{1}{5t}{\sigma }_2=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}{t}^{-1/5}{\dot{\sigma }}_2-\frac{1}{5}{t}^{-6/5}{\sigma }_2=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}\frac{d}{dt}(t^{-1/5}{\sigma }_2)=0.$$

      Integrating and using the condition ${\sigma }_2(15/4)=15/4$ yields

      $$t^{-1/5}{\sigma }_2(t)-{\left(\frac{15}{4}\right)}^{-1/5}{\sigma }_2\left(\frac{15}{4}\right)=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}\boxed{{\sigma }_2(t)={\left(\frac{15}{4}\right)}^{4/5}{t}^{1/5}}$$

      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

      where ${\sigma }_1(t)=t/5+3$, ${\sigma }_2(t)={(\frac{15}{4})}^{\frac{4}{5}}{t}^{\frac{1}{5}}$. This is illustrated in Figure 6.

      

      Figure 6: Characteristics, shocks, and the expansion fan for Q10.

11. Watching paint dry.

    • The conservation law has flux $f(u)=\frac{1}{3}{u}^3$ and $c(u)=f^{\prime }(u)=u^2$. The characteristics are

      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

      $$\dot{\sigma }=\frac{[[f(u)]]}{[[u]]}=\frac{\frac{1}{3}{u}_l^3-\frac{1}{3}{u}_r^3}{u_l-u_r}=\frac{1}{3}.$$

      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

      

      Figure 7: Characteristics for Q11.

      

      Figure 8: Characteristics and the shock $x=\sigma (t)$ for Q11.

      The characteristics tell us nothing about the solution in the region . We can fill the void with lines of the form $x/t=C$, $C\in [0,1]$. These are shown in green in Figure 9.

      

      Figure 9: Characteristics, the shock $x=\sigma (t)$, and the expansion fan for Q11.

      We seek a rarefaction wave that is constant along these lines of the form $u(x,t)=h(x/t)$, $h(0)=0$, $h(1)=1$:

      $$u_t+u^2{u}_x=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}-\frac{x}{t^2}{h}^{\prime }+h^2\frac{1}{t}{h}^{\prime }=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}\frac{h^{\prime }}{t}\left(h^2-\frac{x}{t}\right)=0.$$

      Therefore $u(x,t)=h(x/t)={(x/t)}^{1/2}$ for . We have derived the following weak solution for :

      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

      $${\dot{\sigma }}_2=\frac{[[f(u)]]}{[[u]]}=\frac{\frac{1}{3}{u}_l^3-\frac{1}{3}{u}_r^3}{u_l-u_r}=\frac{{\sigma }_2}{3t}\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}{\dot{\sigma }}_2-\frac{1}{3t}{\sigma }_2=0.$$

      We can solve this ODE using the integrating factor

      $$e^{-\int \frac{1}{3t}𝑑t}=e^{-\frac{1}{3}\ln t}=t^{-1/3}$$

      as follows:

      $$\dot{{\sigma }_2}-\frac{1}{3t}{\sigma }_2=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}{t}^{-1/3}{\dot{\sigma }}_2-\frac{1}{3}{t}^{-4/3}{\sigma }_2=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}\frac{d}{dt}(t^{-1/3}{\sigma }_2)=0.$$

      Integrating and using the condition ${\sigma }_2(3/2)=3/2$ yields

      $$t^{-1/3}{\sigma }_2(t)-{\left(\frac{3}{2}\right)}^{-1/3}{\sigma }_2\left(\frac{3}{2}\right)=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}{\sigma }_2(t)={\left(\frac{3}{2}\right)}^{2/3}{t}^{1/3}.$$

      The shock is easier to sketch if we write it in the form $t=\frac{4}{9}{x}^3$. We have found the weak solution

      where $\sigma (t)=1+t/3$, ${\sigma }_2(t)={(\frac{3}{2})}^{\frac{2}{3}}{t}^{\frac{1}{3}}$. This is illustrated in Figure 10.

      

      Figure 10: Characteristics, shocks, and the expansion fan for Q11.

12. Traffic flow.

    • Let $f(\rho )=\rho v(\rho )=v_m\rho (1-\rho )$. Then

      $$f^{\prime }(\rho )=v_m(1-\rho )-v_m\rho =v_m(1-2\rho )$$

      and therefore

      $${\rho }_t+f{(\rho )}_x=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}{\rho }_t+v_m(1-2\rho ){\rho }_x=0$$

      as required.

    • Let $c(\rho )=f^{\prime }(\rho )=v_m(1-2\rho )$. The characteristics for the conservation law are

      For , the characteristics have slope $1/c({\rho }_0(s))=2/v_m$. The characteristics for 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

      $\dot{{\sigma }_1}(t)$ $={\displaystyle \frac{[[f(\rho )]]}{[[\rho ]]}}$

      $={\displaystyle \frac{v_m{\rho }_l(1-p_l)-v_m{\rho }_r(1-{\rho }_r)}{{\rho }_l-{\rho }_r}}$

      $={\displaystyle \frac{v_m({\rho }_l-{\rho }_r)(1-{\rho }_l-{\rho }_r)}{{\rho }_l-{\rho }_r}}$

      $=v_m(1-{\rho }_l-{\rho }_r)$

      $={\displaystyle \frac{5}{12}}{v}_m.$

      Therefore ${\sigma }_1(t)=\frac{5}{12}{v}_{m}t$. The characteristics and the shock are shown in Figure 11.

      

      Figure 11: Characteristics and the shock $x={\sigma }_1(t)$ for Q12 with $v_m=12$.

      The characteristics tell us nothing about the solution in the void . 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.

      

      Figure 12: Characteristics, the expansion fan, and the shock $x={\sigma }_1(t)$ for Q12 with $v_m=12$.

      We seek a rarefaction wave that is constant along these lines of the form $\rho (x,t)=h((x-1)/t)$ with $h(v_m/3)=1/3$, $h(v_m/2)=1/4$:

      ${\rho }_t+v_m(1-2\rho ){\rho }_x=0$ $⟺\mathit{\hspace{1em}}-{\displaystyle \frac{(x-1)}{t^2}}{h}^{\prime }+v_m(1-2h){\displaystyle \frac{1}{t}}{h}^{\prime }=0$

      $⟺\mathit{\hspace{1em}}{\displaystyle \frac{h^{\prime }}{t}}\left(-{\displaystyle \frac{(x-1)}{t}}+v_m-2v_{m}h\right)=0$

      $⟺\mathit{\hspace{1em}}h\left({\displaystyle \frac{x-1}{t}}\right)={\displaystyle \frac{1}{2}}-{\displaystyle \frac{x-1}{2v_{m}t}}={\displaystyle \frac{v_{m}t-x+1}{2v_{m}t}}.$

      (As a consistency check, note that $h(y)=\frac{1}{2}-\frac{y}{2v_m}$ satisfies $h(v_m/3)=1/3$, $h(v_m/2)=1/4$ as desired.) Therefore

      $$\rho (x,t)=\frac{v_{m}t-x+1}{2v_{m}t}$$

      for , for sufficiently small $t$. After some point in time the rarefaction wave collides with the characteristics with . See Figure 13.

      

      Figure 13: Characteristics, the expansion fan, and the shocks for Q12 with $v_m=12$.

      This collision first occurs when

      $${\sigma }_1(t)=1+\frac{v_m}{3}t\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}t=\frac{12}{v_m}.$$

      The corresponding value of $x$ is

      $$x={\sigma }_1\left(\frac{12}{v_m}\right)=5.$$

      Now a second shock $x={\sigma }_2(t)$ forms with ${\sigma }_2(\frac{12}{v_m})=5$, $\rho =1/4$ on the left of the shock, and $\rho =(v_{m}t-x+1)/(2v_{m}t)$ on the right of the shock. Therefore ${\rho }_l=1/4$, ${\rho }_r=(v_{m}t-{\sigma }_2(t)+1)/(2v_{m}t)$, 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

      $$\frac{d}{dt}(t^{-1/2}{\sigma }_2(t))=\frac{v_m}{4}{t}^{-1/2}-\frac{1}{2}{t}^{-3/2}.$$

      Now integrate and use the initial condition ${\sigma }_2(\frac{12}{v_m})=5$ to obtain

      $t^{-1/2}{\sigma }_2(t)-{\left(\frac{12}{v_m}\right)}^{-1/2}{\sigma }_2\left(\frac{12}{v_m}\right)$ $={\displaystyle {\int }_{12/v_m}^t}\left({\displaystyle \frac{v_m}{4}}{s}^{-1/2}-{\displaystyle \frac{1}{2}}{s}^{-3/2}\right)𝑑s$

      $={\displaystyle \frac{v_m}{2}}\left[t^{1/2}-{\left(\frac{12}{v_m}\right)}^{1/2}\right]+t^{-1/2}-{\left(\frac{12}{v_m}\right)}^{-1/2}.$

      Therefore

      $${\sigma }_2(t)=\frac{v_{m}t}{2}+1+\left(4{\left(\frac{12}{v_m}\right)}^{-1/2}-\frac{v_m}{2}{\left(\frac{12}{v_m}\right)}^{1/2}\right)t^{1/2}=\frac{v_{m}t}{2}+1-{\left(\frac{v_{m}t}{3}\right)}^{1/2}.$$

      We have arrived at the following weak solution:

      where

      $$\boxed{{\sigma }_1(t)=\frac{5v_{m}t}{12},{\sigma }_2(t)=\frac{v_{m}t}{2}+1-{\left(\frac{v_{m}t}{3}\right)}^{1/2}}$$

    • 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.

      

      Figure 14: The weak solution found in Q12 at times $t=0$, $6/v_m$, $24/v_m$ with $v_m=12$.

13. 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 $V=ℝ\times (0,\mathrm{\infty })$ and let $\phi \in C_c^{\mathrm{\infty }}(V)$. Then

    ${\displaystyle {\int }_V}[u{\phi }_t+f(u){\phi }_x]𝑑x𝑑t+{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{u}_0(x)\phi (x,0)𝑑x$

    $={\displaystyle {\int }_{V_l}}[u{\phi }_t+f(u){\phi }_x]𝑑x𝑑t+{\displaystyle {\int }_{V_r}}[u{\phi }_t+f(u){\phi }_x]𝑑x𝑑t+{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{u}_0(x)\phi (x,0)𝑑x$

    $={\displaystyle {\int }_{V_l}}(f(u),u)\cdot {\nabla }_{(x,t)}\phi dxdt+{\displaystyle {\int }_{V_r}}(f(u),u)\cdot {\nabla }_{(x,t)}\phi dxdt+{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{u}_0(x)\phi (x,0)𝑑x$

    $={\displaystyle {\int }_{\partial V_l}}(f(u),u)\phi \cdot {𝒏}_l𝑑L-{\displaystyle {\int }_{V_l}}{\mathrm{div}}_{(x,t)}(f(u),u)\phi 𝑑x𝑑t+{\displaystyle {\int }_{\partial V_r}}(f(u),u)\phi \cdot {𝒏}_r𝑑L-{\displaystyle {\int }_{V_r}}{\mathrm{div}}_{(x,t)}(f(u),u)\phi 𝑑x𝑑t$

    $+{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{u}_0(x)\phi (x,0)𝑑x$

    where ${𝒏}_l$ and ${𝒏}_r$ are the unit outward-pointing normal vectors to $\partial V_l$, $\partial V_r$. Observe that

    $$\partial V_l=\{(x,0):x\le g(0)\}\cup \gamma ,\partial V_r=\{(x,0):x\ge g(0)\}\cup \gamma .$$

    Moreover ${𝒏}_r=-{𝒏}_l\text{ on }\gamma$, ${𝒏}_l=-(0,1)$ on $\partial V_l\setminus \gamma$, ${𝒏}_r=-(0,1)$ on $\partial V_r\setminus \gamma$. Continuing the calculation gives

    ${\displaystyle {\int }_V}[u{\phi }_t+f(u){\phi }_x]𝑑x𝑑t+{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{u}_0(x)\phi (x,0)𝑑x$

    $=-{{\displaystyle {\int }_{-\mathrm{\infty }}^{g(0)}}(f(u),u)\phi |}_{t=0}\cdot (0,1)dx+{\displaystyle {\int }_{\gamma }}(f(u),u)\phi \cdot {𝒏}_l𝑑L-{\displaystyle {\int }_{V_l}}(\underset{=0}{\underbrace{u_t+f{(u)}_x}})\phi 𝑑x𝑑t$

    $-{{\displaystyle {\int }_{g(0)}^{\mathrm{\infty }}}(f(u),u)\phi |}_{t=0}\cdot (0,1)dx+{\displaystyle {\int }_{\gamma }}(f(u),u)\phi \cdot {𝒏}_r𝑑L-{\displaystyle {\int }_{V_r}}(\underset{=0}{\underbrace{u_t+f{(u)}_x}})\phi 𝑑x𝑑t+{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{u}_0(x)\phi (x,0)𝑑x$

    $=-{\displaystyle {\int }_{-\mathrm{\infty }}^{g(0)}}u(x,0)\phi (x,0)𝑑x+{\displaystyle {\int }_{\gamma }}(f(u),u)\phi \cdot {𝒏}_l𝑑L-{\displaystyle {\int }_{V_l}}(\underset{=0}{\underbrace{u_t+f{(u)}_x}})\phi 𝑑x𝑑t$

    $-{\displaystyle {\int }_{g(0)}^{\mathrm{\infty }}}u(x,0)\phi (x,0)𝑑x+{\displaystyle {\int }_{\gamma }}(f(u),u)\phi \cdot \underset{=-{𝒏}_l}{\underbrace{{𝒏}_r}}𝑑L-{\displaystyle {\int }_{V_r}}(\underset{=0}{\underbrace{u_t+f{(u)}_x}})\phi 𝑑x𝑑t+{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{u}_0(x)\phi (x,0)𝑑x$

    $={\displaystyle {\int }_{\mathrm{\infty }}^{\mathrm{\infty }}}(\underset{=0}{\underbrace{u_0(x)-u(x,0)}})\phi (x,0)𝑑x$

    $=0.$

    This holds for all $\phi \in C_c^{\mathrm{\infty }}(V)$. Therefore $u$ is a weak solution of the conservation law, as required.