[problem: Burgers Equation With Piecewise Linear Data | 2] Consider the Cauchy problem for the Burgers equation: \begin{align*} \partial_t u + u\,\partial_x u &= 0, \quad x \in \mathbb{R},\; t > 0, \\ u(0, x) &= g(x), \quad x \in \mathbb{R}, \end{align*} where \begin{align*} g(x) = \begin{cases} 0, & x \in (-\infty, -1], \\ -x, & x \in (-1, 0], \\ 0, & x \in (0, \infty). \end{cases} \end{align*} (a) Find the unique entropy solution of the problem in the time interval $t \in (0, 1)$. (b) Find the unique entropy solution of the problem in the time interval $t \in (1, \infty)$. [/problem]

[solution] **Step 1: Identify the structure of the initial data.** The initial datum $g$ is piecewise linear with three regions. At $x = 0$, $g$ is continuous: $g(0^-) = 0 = g(0^+)$. At $x = -1$, $g$ is discontinuous: $g(-1^-) = 0$ but $\lim_{x \to -1^+} g(x) = 1$. This upward jump from $u_L = 0$ to $u_R = 1$ and the subsequent decrease from $1$ to $0$ on $(-1, 0)$ are the two features that determine the solution structure. **Step 2: Compute characteristics from each region.** Writing $(x_1, x_2) = (x, t)$ so that $n = 2$, the Burgers equation corresponds to $F(a, b, c) = a_2 + b\,a_1$. The partial derivatives are $\partial F / \partial a_1 = b$, $\partial F / \partial a_2 = 1$, $\partial F / \partial b = a_1$, and $\partial F / \partial c_i = 0$. The characteristic ODE system therefore reads: \begin{align*} \dot{X}_1(s) &= U(s), \quad \dot{X}_2(s) = 1, \\ \dot{G}_1(s) &= -(G_1(s))^2, \quad \dot{G}_2(s) = -G_2(s)\,G_1(s), \\ \dot{U}(s) &= G_1(s)\,U(s) + G_2(s). \end{align*} Since $F$ is preserved along characteristics and $F(T(0)) = 0$, we have $G_2(s) + U(s)\,G_1(s) = 0$ for all $s$, and therefore $\dot{U}(s) = 0$. From the initial data $X_1(0) = x_0$, $X_2(0) = 0$, $U(0) = g(x_0)$, integrating gives: \begin{align*} U(s) = g(x_0), \quad X_1(s) = x_0 + s\,g(x_0), \quad X_2(s) = s. \end{align*} We now apply this to each region of $g$. *Region 1: $x_0 \leq -1$.* Here $g(x_0) = 0$, so the parameterised characteristic curve is \begin{align*} (X_1(s),\, X_2(s)) = (x_0,\, s), \quad s \geq 0, \end{align*} carrying the constant value $U(s) = 0$. Geometrically, this is the vertical ray in the $(x,t)$-plane issuing from $(x_0, 0)$. *Region 2: $x_0 \in (-1, 0)$.* Here $g(x_0) = -x_0 > 0$, so the parameterised characteristic curve is \begin{align*} (X_1(s),\, X_2(s)) = \bigl(x_0(1 - s),\, s\bigr), \quad s \geq 0, \end{align*} carrying the constant value $U(s) = -x_0$. This is a straight line in the $(x,t)$-plane passing through $(x_0, 0)$ with slope $dt/dx = 1/g(x_0) = -1/x_0$. As $x_0$ varies over $(-1, 0)$, these lines fan out from the point $(-1, 0)$: the characteristic from $x_0 = -1 + \epsilon$ has speed close to $1$, while the characteristic from $x_0 = -\epsilon$ has speed close to $0$. *Region 3: $x_0 > 0$.* Here $g(x_0) = 0$, so the parameterised characteristic curve is \begin{align*} (X_1(s),\, X_2(s)) = (x_0,\, s), \quad s \geq 0, \end{align*} carrying the constant value $U(s) = 0$. As in Region 1, these are vertical rays. **Strategy for recovering the solution.** To find $u(x,t)$ at a given point $(x,t)$ in the $(x,t)$-plane, we must determine which characteristic passes through it — that is, find $x_0$ and $s$ such that $(X_1(s), X_2(s)) = (x, t)$. Since $X_2(s) = s$ in all three regions, we immediately obtain $s = t$. It remains to invert the relation $X_1(t) = x$ for $x_0$: if this can be done uniquely, the solution is $u(x,t) = U(t) = g(x_0)$, and we have a classical solution in that region of the $(x,t)$-plane. If multiple characteristics pass through the same point (crossing), or if no characteristic reaches a given point (a gap), the method of characteristics alone does not determine the solution there, and we must appeal to weak solution theory (shocks or rarefaction waves respectively). We now check whether crossing or gaps occur for $t < 1$. **Step 3: Check for crossing among Region 2 characteristics ($t < 1$).** Within Region 2, the map $x_0 \mapsto X_1(t; x_0) = x_0(1-t)$ has derivative $(1-t)$ with respect to $x_0$. For $t < 1$ this is strictly positive, so the map is injective — no two Region 2 characteristics meet before $t = 1$. As $x_0$ ranges over $(-1, 0)$, the position $x_0(1-t)$ ranges over $(-(1-t), 0)$. The Region 2 characteristics do not enter $x > 0$ (since $x_0(1-t) < 0$ for $x_0 < 0$, $t < 1$), so they do not cross Region 3 characteristics either. At $t = 1$, every Region 2 characteristic arrives at $x_0(1-1) = 0$: all characteristics converge to the single point $(x, t) = (0, 1)$. **Step 4: Resolve the discontinuity at $x = -1$ (rarefaction wave).** At $x = -1$, the left state is $u_L = g(-1^-) = 0$ and the right state is $u_R = \lim_{x \to -1^+} g(x) = 1$. Since $u_L < u_R$, the characteristics fan apart: the left characteristic (from Region 1) is vertical ($\dot{X}_1 = 0$), while the rightmost Region 2 characteristic has speed close to $1$. This leaves a wedge-shaped gap $\{(x, t) : -1 < x < -1 + t\}$ that no characteristic from either side reaches. Since the flux $q(u) = u^2/2$ is convex, the entropy condition selects a rarefaction wave filling this gap. The self-similar rarefaction centred at $x = -1$ connecting $u_L = 0$ to $u_R = 1$ is found by seeking a solution of the form $u = \phi((x+1)/t)$. Substituting into $\partial_t u + u\,\partial_x u = 0$ gives $\phi(\xi) = \xi$ where $\xi = (x+1)/t$, so: \begin{align*} u(x,t) = \frac{x + 1}{t}, \quad -1 \leq x \leq -1 + t. \end{align*} One verifies: $\partial_t u = -(x+1)/t^2$ and $\partial_x u = 1/t$, so $\partial_t u + u\,\partial_x u = -(x+1)/t^2 + (x+1)/t^2 = 0$. At the boundaries: $u = 0$ at $x = -1$ (matching Region 1) and $u = 1$ at $x = -1 + t$ (matching Region 2, see below). **Step 5: Extract the explicit solution from Region 2.** For $x_0 \in (-1, 0)$, the characteristic triple gives $U(s) = u(X_1(s), X_2(s))$. Substituting the expressions from Step 2: \begin{align*} u\bigl(x_0(1 - s),\; s\bigr) = -x_0 \quad \text{for all } s \in [0, 1). \end{align*} Writing $x = X_1(s) = x_0(1 - s)$ and $t = X_2(s) = s$, we solve for $x_0$: since $x = x_0(1-t)$ and $t < 1$, we obtain $x_0 = x/(1-t)$. The solution value is $u = -x_0 = -x/(1-t)$, giving the explicit formula: \begin{align*} u(x,t) = \frac{-x}{1 - t}, \quad x \in (-(1-t),\, 0),\; t \in (0,1). \end{align*} We verify that this candidate solves the PDE: $\partial_t u = -x/(1-t)^2$ and $\partial_x u = -1/(1-t)$, so \begin{align*} \partial_t u + u\,\partial_x u = \frac{-x}{(1-t)^2} + \frac{-x}{1-t} \cdot \frac{-1}{1-t} = \frac{-x}{(1-t)^2} + \frac{x}{(1-t)^2} = 0. \quad \checkmark \end{align*} Checking continuity at the boundaries: At $x = -(1-t)$: the rarefaction gives $u = (-(1-t)+1)/t = t/t = 1$; Region 2 gives $u = (1-t)/(1-t) = 1$. ✓ At $x = 0$: Region 2 gives $u = 0/(1-t) = 0$; Region 3 gives $u = 0$. ✓ **Step 6: Assemble the entropy solution for $t \in (0,1)$ (part (a)).** \begin{align*} u(x,t) = \begin{cases} 0, & x < -1, \\ \dfrac{x+1}{t}, & -1 \leq x \leq -1+t, \\[6pt] \dfrac{-x}{1-t}, & -1+t \leq x \leq 0, \\[4pt] 0, & x > 0. \end{cases} \end{align*} **Step 7: Analyse what happens at $t = 1$ and form the shock (part (b)).** At $t = 1$, all Region 2 characteristics converge to $x = 0$. For $t > 1$ the Region 2 solution $u = -x/(1-t)$ ceases to exist. The rarefaction wave now carries values up to $u_L = (\sigma(t)+1)/t > 0$ at its right edge, which meets the region $u_R = 0$ to the right. Since $u_L > u_R$, a shock forms at $x = 0$ at time $t = 1$. **Step 8: Track the shock curve for $t > 1$ using Rankine–Hugoniot.** The shock separates $u_L(t)$ on the left (the value at the right edge of the rarefaction) from $u_R = 0$ on the right. If the shock is at position $x = \sigma(t)$, then the right edge of the rarefaction is at $x = \sigma(t)$, so $u_L = (\sigma(t) + 1)/t$. The Rankine–Hugoniot condition for $q(u) = u^2/2$ gives the shock speed: \begin{align*} \dot{\sigma}(t) = \frac{q(u_L) - q(u_R)}{u_L - u_R} = \frac{u_L^2/2}{u_L} = \frac{u_L}{2} = \frac{\sigma(t) + 1}{2t}. \end{align*} This is a first-order linear ODE with initial condition $\sigma(1) = 0$. **Step 9: Solve the shock ODE.** Writing $v(t) = \sigma(t) + 1$, the equation becomes $\dot{v} = v/(2t)$, which is separable: \begin{align*} \frac{dv}{v} = \frac{dt}{2t} \quad \implies \quad \ln|v| = \tfrac{1}{2}\ln t + C \quad \implies \quad v = A\sqrt{t}. \end{align*} From $\sigma(1) = 0$: $v(1) = 1$, so $A = 1$. Therefore $\sigma(t) = \sqrt{t} - 1$. One checks the entropy condition: $u_L = (\sqrt{t} - 1 + 1)/t = 1/\sqrt{t} > 0 = u_R$, so the Lax entropy condition $u_L > u_R$ holds for convex flux. ✓ **Step 10: Assemble the entropy solution for $t > 1$ (part (b)).** \begin{align*} u(x,t) = \begin{cases} 0, & x < -1, \\ \dfrac{x+1}{t}, & -1 \leq x \leq \sqrt{t} - 1, \\[6pt] 0, & x > \sqrt{t} - 1. \end{cases} \end{align*} [/solution]